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In this video, we will learn what is meant by the critical points of a function.
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We’ll learn about the types of critical points that exist and how to find the critical points of a function using differentiation.
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We’ll also see how to apply the first-derivative test in order to classify critical points.
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Firstly then, what are critical points?
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They’re sometimes called stationary or turning points.
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And they’re really important features of the graph of a function.
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They’re points at which the gradient of the graph — that’s d𝑦 by d𝑥 — is equal to zero or is undefined.
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If the function was specified as 𝑦 equals 𝑓 of 𝑥, then its points when 𝑓 prime of 𝑥 is equal to zero.
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There are three types of critical points that we need to be aware of.
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The first are local maxima, which are points where the value of the function is the highest in a neighbouring region around that point.
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These are distinguished by the gradient of the graph being positive on the left of the maximum point.
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That’s for 𝑥-values less than the 𝑥-value at the maximum.
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And being negative to the right of the maximum point.
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That’s for 𝑥-values greater than the 𝑥-value at the maximum.
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An example of this would be in the graph of 𝑦 equals negative three 𝑥 squared.
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The second type of critical points that we need to be aware of are local minima, which are points where the value of the function is the lowest than it is in a neighbouring region.
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These are characterised by having a negative gradient to the left and a positive gradient to the right, such as the turning point on the graph of 𝑦 equals 𝑥 squared.
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The final type of critical point are points of inflection.
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And these are characterised by the gradient having the same sign on either side of the critical point.
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So it could be positive on both sides, such as in the graph of 𝑦 equals 𝑥 cubed.
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Or it could be negative on both sides, such as in the graph of 𝑦 equals negative 𝑥 cubed.
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Remember that, at each turning point itself, the gradient d𝑦 by d𝑥 will be equal to zero.
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To find a critical point, we therefore first need to find the gradient function for the curve d𝑦 by d𝑥.
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Once we found this, we can set d𝑦 by d𝑥 equal to zero and solve the resulting equation in order to find the 𝑥-coordinate of the critical point.
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Usually, we also want to know the 𝑦-coordinate to the critical point or the value of the function, which we can find by substituting our 𝑥-value or values back into the equation of the curve.
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We’ll look at some examples of this.
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And we’ll also discuss one method of determining the type of critical point that we have.
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Determine the critical points of the function 𝑦 equals negative eight 𝑥 cubed in the interval negative two, one.
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First, we recall that, at the critical points of a function, the gradient d𝑦 by d𝑥 will be equal to zero.
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So we need to find the gradient function for this curve.
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We can apply the power rule of differentiation.
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And it tells us that d𝑦 by d𝑥 is equal to negative eight multiplied by three multiplied by 𝑥 squared, which simplifies to negative 24𝑥 squared.
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Next, we set our expression for d𝑦 by d𝑥 equal to zero, giving the equation negative 24𝑥 squared equals zero.
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We now need to solve for 𝑥.
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And we can see that as negative 24 isn’t equal to zero, it must be the case that 𝑥 squared is equal to zero.
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And if 𝑥 squared is equal to zero, then 𝑥 itself must be equal to zero.
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So we found the value of 𝑥 at our critical point.
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Now in the question, we were asked to determine the critical points only in a particular interval, the interval negative two to one.
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And our value of 𝑥 does lie in this interval.
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In fact, it is the only critical point of this function.
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Next, we need to find the value of 𝑦 at this critical point, which we can do by substituting our 𝑥-value back into the equation of the function.
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The function was 𝑦 equals negative eight 𝑥 cubed.
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So we have 𝑦 equals negative eight multiplied by zero cubed, which is just zero.
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So the only critical point in this interval and in fact the only critical point for the entire function has coordinates zero, zero.
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We can also see this if we sketch a graph of 𝑦 equals negative eight 𝑥 cubed.
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Now this is the graph of 𝑦 equals 𝑥 cubed, which we should be familiar with.
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And it has a critical point.
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In fact, it’s a point of inflection at the origin.
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Multiplying by eight will cause a stretch with a scale factor of eight in the vertical direction.
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But this doesn’t affect the critical point.
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And then multiplying by negative one will cause a reflection in the 𝑥-axis.
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So in green, we have the graph of 𝑦 equals negative eight 𝑥 cubed.
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We can see that it does indeed have a critical point a point of inflection at the origin.
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So by first finding the gradient function d𝑦 by d𝑥 and then setting it equal to zero and solving the resulting equation, we found the 𝑥-coordinate of our critical point.
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We then substituted this back into the equation of the original function in order to find the corresponding 𝑦-coordinate, giving a critical point of zero, zero.
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In our next example, we’ll consider how to determine the type of critical point without needing to sketch a graph.
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Determine, if any, the local maximum and minimum values of 𝑓 of 𝑥 equals negative two 𝑥 cubed minus nine 𝑥 squared minus 12𝑥 minus 15 together with where they occur.
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In this question, we were asked to determine the local maximum and minimum values.
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So that’s the values of the function itself.
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And “with where they occur” means we also need to find the corresponding 𝑥-values.
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First, we recall that, at the critical points of a function, the gradient — so that’s 𝑓 prime of 𝑥 — is equal to zero.
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So we begin by differentiating 𝑓 of 𝑥, which we can do using the power rule, to give negative six 𝑥 squared minus 18𝑥 minus 12.
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At critical points, 𝑓 prime of 𝑥 is equal to zero.
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So we set our expression for 𝑓 prime of 𝑥 equal to zero.
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And we’ll now solve the resulting equation for 𝑥.
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We can factor negative six from this equation, giving negative six multiplied by 𝑥 squared plus three 𝑥 plus two is equal to zero.
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And we see that, inside the parentheses, we have a quadratic in 𝑥 which does factorise.
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It’s equal to 𝑥 plus two multiplied by 𝑥 plus one.
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We also note at this point that negative six is not equal to zero.
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So we can eliminate it from our equation at this point.
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We now set each factor in turn equal to zero, giving 𝑥 plus two equals zero or 𝑥 plus one equals zero.
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Both equations can be solved in a relatively straightforward way to give the 𝑥-coordinates of the critical points of this function.
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𝑥 is equal to negative two or 𝑥 is equal to negative one.
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We now know the 𝑥-values at our critical points.
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But we also need to know the values of the function itself.
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So we need to evaluate 𝑓 of 𝑥 at each critical value.
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When 𝑥 is equal to negative two, 𝑓 of negative two is negative two multiplied by negative two cubed minus nine multiplied by negative two squared minus 12 multiplied by negative two minus 15, which is equal to negative 11.
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Evaluating 𝑓 of negative one in the same way gives negative 10.
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So we’ve now established that this function has critical points at negative two, negative 11 and negative one, negative 10.
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But how do we determine whether they’re local maxima or minima or indeed points of inflection?
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Well, we’re going to use something called the first-derivative test.
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We’ll consider the sign of the derivative either side of our critical point, which will tell us the gradient of the function either side of the critical point.
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By considering this, we’ll be able to identify the shape of the function near to each critical point.
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Here’s what we’re going to do.
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We’re going to evaluate the first derivative — that’s 𝑓 prime of 𝑥 — a little bit either side of our critical 𝑥-values of negative two and negative one.
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Now, usually, we try to just use the nearest integer values.
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But in this case, negative two and negative one are consecutive integers.
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So instead, we’ve chosen a value between them to be the upper value for negative two and the lower value for negative one.
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We’ve chosen a value of negative 1.5.
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Remember that our gradient function 𝑓 prime of 𝑥 is negative six 𝑥 squared minus 18𝑥 minus 12.
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So when we evaluate this at negative three, we get a value of negative 12.
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Now we’re not particularly interested in what the value is, but rather its sign.
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So negative 12 is a negative value.
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We’ll also evaluate our gradient function at negative 1.5.
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And it gives 1.5, which is a positive value.
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Finally, we need to evaluate this gradient function at zero.
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And it gives negative 12, a negative value.
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So how does this help us with determining whether the critical points are maxima or minima?
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Well, we see that the gradient of this curve is negative when 𝑥 is equal to negative three.
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It’s then equal to zero when 𝑥 equals negative two, and it’s positive when 𝑥 equals negative 1.5.
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And by sketching that shape, we see that the critical point at negative two must be a local minimum.
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In the same way, the gradient of this function is positive when 𝑥 equals negative 1.5.
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It’s zero when 𝑥 equals negative one.
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And it’s negative when 𝑥 equals zero.
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So we see that the critical point at 𝑥 equals negative one must be a local maximum.
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So this method, the first-derivative test, we consider the first derivative — that’s the gradient — either side of the critical point.
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And by considering the sign of this gradient, we can deduce the shape of the curve at this point.
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We found then that this function, 𝑓 of 𝑥, has a local minimum at negative two, negative 11 and a local maximum at negative one, negative 10.
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There is also another method for determining the nature of critical points, called the second-derivative test.
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It involves differentiating the gradient function to give the second derivative of the original function.
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As the first derivative reveals how the function itself is changing, the second derivative reveals how the gradient is changing.
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And so by considering this, we can determine whether a point is a local minimum, a local maximum, or a point of inflection.
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However, this is out of the scope of what we’re going to look at in this video.
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Given that the function 𝑓 of 𝑥 equals 𝑥 squared plus 𝐿 𝑥 plus 𝑀 has a minimum value of two at 𝑥 equals negative one, determine the values of 𝐿 and 𝑀.
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In this question, we’ve been told the minimum value of the function.
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It is two.
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And we’ve been told the value of 𝑥 at which this occurs.
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It’s negative one.
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We need to use this information to calculate the missing coefficients 𝐿 and 𝑀 in the definition of 𝑓 of 𝑥.
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A minimum is a type of critical point.
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And we recall then that, at critical points, the gradient of the function 𝑓 prime of 𝑥 is equal to zero.
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We can use the power rule to differentiate 𝑓 of 𝑥.
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And we have that 𝑓 prime of 𝑥 is equal to two 𝑥 plus 𝐿.
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As negative one is the 𝑥-value at a critical point, we know that if we substitute 𝑥 equals negative one into our expression for 𝑓 prime of 𝑥, we must get a result of zero.
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So we can form an equation.
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Two multiplied by negative one plus 𝐿 equals zero.
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This gives the equation negative two plus 𝐿 equals zero, which we can solve to give 𝐿 equals two.
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So we found the value of 𝐿.
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But what about the value of 𝑀?
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Well, we know that the function has a minimum value of two when 𝑥 equals negative one.
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So when 𝑥 equals negative one, 𝑓 of 𝑥 is equal to two.
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So we can substitute negative one for 𝑥, two for 𝐿, and two for 𝑓 of 𝑥 to give a second equation.
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Negative one squared plus two multiplied by negative one plus 𝑀 is equal to two.
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This simplifies to one minus two plus 𝑀 is equal to two, which again we can solve to give 𝑀 is equal to three.
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We found the values of 𝐿 and 𝑀.
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𝐿 is equal to two, and 𝑀 is equal to three.
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But let’s just confirm that this point is indeed a minimum point.
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We can do this using the first-derivative test.
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We evaluate the first derivative, 𝑓 prime of 𝑥, either side of our critical value of 𝑥 so that we can see how the gradient of this function is changing around the critical point.
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Remember, at the critical point itself, the gradient is equal to zero.
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Our gradient function 𝑓 prime of 𝑥 was two 𝑥 plus 𝐿.
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But we now know that 𝐿 is equal to two.
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So our gradient function is two 𝑥 plus two.
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When 𝑥 is equal to negative two, this will give negative two.
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And when 𝑥 is equal to zero, this will give positive two.
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It’s actually the sign of the value rather than the value itself that we’re interested in.
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We see that the gradient is negative to the left of negative one.
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It’s then zero at negative one itself and then positive to the right of negative one.
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So by sketching this pattern, we see that the critical point at negative one is indeed a local minimum.
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We finished the problem then.
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𝐿 is equal to two, and 𝑀 is equal to three.
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Sometimes, the functions we differentiate will be more complex than polynomials, such as exponential or trigonometric functions.
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We may also need to use one of our differentiation rules, such as the product, quotient or chain rules.
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We’ll see this in our next example.
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Determine where 𝑓 of 𝑥 equals three 𝑥 squared 𝑒 to the power of negative 𝑥 has a local maximum and give the value there.
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We’ll recall first of all that a local maximum is a type of critical point.
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And at critical points, 𝑓 prime of 𝑥 is equal to zero.
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So we need to begin by finding the derivative of 𝑓 of 𝑥.
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Now looking at 𝑓 of 𝑥, we can see that it is actually a product.
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It’s equal to one function, three 𝑥 squared, multiplied by another function, 𝑒 to the power of negative 𝑥.
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So in order to differentiate 𝑓 of 𝑥, we’re going to need to apply the product rule.
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The product rule tells us that the derivative of the product 𝑢𝑣 is equal to 𝑢𝑣 prime plus 𝑢 prime 𝑣.
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We can define 𝑢 to be the function three 𝑥 squared.
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And we can define 𝑣 to be the function 𝑒 to the power of negative 𝑥.
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We now need to differentiate each of these functions.
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We can apply the power rule to differentiate 𝑢.
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And it gives six 𝑥.
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And in order to differentiate 𝑣, we need to recall our rules for differentiating exponentials.
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The derivative with respect to 𝑥 of 𝑒 to the power of 𝑘𝑥 is 𝑘𝑒 to the power of 𝑘𝑥.
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So the derivative of 𝑒 to the power of negative 𝑥 is negative 𝑒 to the power of negative 𝑥.
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The value of 𝑘 here is negative one.
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Applying the product rule then, 𝑓 prime of 𝑥 is equal to 𝑢𝑣 prime — that’s three 𝑥 squared multiplied by negative 𝑒 to the negative 𝑥 — plus 𝑢 prime 𝑣 — that’s six 𝑥 multiplied by 𝑒 to the power of negative 𝑥.
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We can factor by three 𝑥𝑒 to the power of negative 𝑥, leaving three 𝑥𝑒 to the power of negative 𝑥 multiplied by negative 𝑥 plus two.
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Now we set 𝑓 prime of 𝑥 equal to zero, giving three 𝑥𝑒 to the negative 𝑥 times negative 𝑥 plus two is equal to zero.
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Three is not equal to zero, so we can eliminate it from our equation.
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You can think of this as dividing both sides by three.
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We’re then left with 𝑥 is equal to zero or 𝑒 to the power of negative 𝑥 is equal to zero or negative 𝑥 plus two is equal to zero.
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The first and last equations yield solutions for 𝑥 straight away.
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But what about the middle equation?
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Well, actually, there is no solution to this equation.
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If you think of the graph of 𝑒 to the power of negative 𝑥, then the 𝑥-axis is an asymptote.
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There’s no value of 𝑥 for which 𝑒 to the power of negative 𝑥 is equal to zero.
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So the only values of 𝑥 are zero and two.
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We then evaluate 𝑓 of 𝑥 at each of these points, giving zero and 12 over 𝑒 squared.
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Finally, we need to confirm which of these points is a local maximum.
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And we can do this by applying the first-derivative test.
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We evaluate 𝑓 prime of 𝑥 at integer values either side of our two critical 𝑥-values of zero and two.
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And the important thing to note is the sign of these values.
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We see that, for our critical point when 𝑥 equals two, the gradient changes from positive to negative, which means that this is the local maximum point.
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So we conclude that there is a local maximum of 12 over 𝑒 squared when 𝑥 is equal to two.
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Let’s summarise then what we’ve learned.
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At the critical points of a function, d𝑦 by d𝑥 or 𝑓 prime of 𝑥 is equal to zero or it’s undefined.
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The three types of critical points that we need to be aware of are local maxima, local minima, and points of inflection.
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We can use the first-derivative test to consider the gradient either side of a critical point and hence classify it as either a local maximum, local minimum, or point of inflection.