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Find, if any, of the local maximum and local minimum values of the function π¦ equal seven π₯ plus seven over π₯.
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Firstly, we recall that local maxima and local minima are examples of critical points of a function.
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And the critical points of a function occur when its first derivative, in this case, dπ¦ by dπ₯, is either equal to zero or does not exist.
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Itβs undefined.
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We, therefore, need to find an expression for the first derivative of this function.
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We will find it helpful, though, to rewrite our function π¦ slightly to begin with, or, more specifically, the second term.
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We can use laws of exponents to rewrite seven over π₯ as seven multiplied by π₯ to the power of negative one.
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We can then see that our function π¦ is the sum of two general power terms.
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And so, we can recall the power rule of differentiation, which tells us that the derivative with respect to π₯ of ππ₯ to the πth power for real values of π and π is equal to ππ multiplied by π₯ to the π minus oneth power.
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We multiply by the original exponent π and then reduce the exponent by one.
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Now, depending how much practice youβve had of differentiation, you may find it helpful to think of seven π₯ as seven π₯ to the first power.
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And then, we can apply the power rule of differentiation.
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Differentiating seven π₯ to the first power gives seven multiplied by one multiplied by π₯ to the zeroth power, which weβll simplify in a moment.
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And then, differentiating seven π₯ to the power of negative one gives seven multiplied by negative one multiplied by π₯ to the power of negative two.
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Now, recall that π₯ to the power of zero is just one.
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So, our first term can be rewritten as seven.
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Itβs just seven multiplied by one.
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And then, to the second term, we can rewrite this as negative seven over π₯ squared.
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So, we have our expression for the first derivative dπ¦ by dπ₯.
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To find the π₯-values at any critical points, we then set this expression for dπ¦ by dπ₯ equal to zero, giving seven minus seven over π₯ squared equals zero.
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We can multiply through by π₯ squared in order to eliminate the denominator in the fraction, giving seven π₯ squared minus seven equals zero.
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We then divide through by seven and add one to each side of the equation to give π₯ squared is equal to one.
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Finally, we need to take the square root of each side of the equation, remembering that we need plus or minus the square root.
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So, we have π₯ is equal to plus or minus the square root of one, which is simply positive or negative one.
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Weβve found then that there were two π₯-values at which our function has critical points.
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We also need to determine the value of the function itself for each of these π₯-values.
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Substituting into our original function π¦ then, which was seven π₯ plus seven over π₯, we have that when π₯ is equal to one, π¦ is equal to seven multiplied by one plus seven over one.
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Thatβs seven plus seven, which is equal to 14.
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And when π₯ is equal to negative one, we find that π¦ is equal to seven multiplied by negative one plus seven over negative one.
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Thatβs negative seven plus negative seven, which is equal to negative 14.
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Weβve found then that the two critical points of this function are at the point with coordinates negative one, negative 14 and the point with coordinates one, 14.
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Finally, we need to determine whether these points are local minima or local maxima or perhaps points of inflection.
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And to do this, weβll apply the second derivative test.
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To do this, we first find a general expression for the second derivative d two π¦ by dπ₯ squared, which we find by differentiating our first derivative again.
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Now, we may find it helpful to think of our first derivative as seven minus seven π₯ to the power of negative two.
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And then, we can apply the power rule of differentiation.
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Again, depending how much practice youβve had with differentiation, you may find it helpful to think of seven as seven π₯ to the power of zero.
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So when we apply the power rule of differentiation, we get seven multiplied by zero multiplied by π₯ to the negative one for the derivative of our first term.
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And then, we have negative seven multiplied by negative two multiplied by π₯ to the power of negative three.
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Of course, multiplying by zero just gives zero.
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So, this entire first term is zero.
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And we remember that the derivative of any constant is always zero.
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Our expression for the second derivative, therefore, simplifies to 14π₯ to the power of negative three, which we can write us 14 over π₯ cubed.
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Finally, we need to evaluate the second derivative at each of our critical points.
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Substituting π₯ equals one, we find that at this point, d two π¦ by dπ₯ squared is equal to 14 over one cubed, which is equal to 14.
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Now, the key point here is that this is a positive value.
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And the second derivative test tells us that if the second derivative is positive at a critical point, then that critical point is a local minimum.
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So, weβve found that the critical point one, 14 is a local minimum of our function π¦.
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At our other critical point, when π₯ is equal to negative one, we find that the second derivative of our function is equal to negative 14.
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So, this time, the second derivative is negative.
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And by the second derivative test, we find that this critical point is a local maximum.
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So, by first finding the critical points of our function and then applying the second derivative test, we found that the local minimum value of the function π¦ is 14 and the local maximum value is negative 14.
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Remember that these local minimum and local maximum values are the values of the function itself at these critical points.
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Although coincidentally, theyβre also the values of the second derivative at these critical points.
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But that is a coincidence.
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Itβs the values of the function itself that weβre interested in.
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Now, just one final point briefly, we said that critical points would occur when the first derivative is either zero or is undefined.
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And we havenβt mentioned anything about where this first derivative might be undefined.
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Well, if you recall, our first derivative was equal to seven minus seven over π₯ squared.
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And this will be undefined when the denominator of that fraction is zero, which will occur when π₯ is equal to zero.
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However, if we look at our original function π¦, we see that π₯ equals zero would also lead to the original function being undefined.
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And therefore, π₯ equals zero is not included in the domain of our function π¦.
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We, therefore, donβt need to be concerned that the first derivative does not exist at this point.
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In summary then, we found that the local minimum value of our function is 14 and the local maximum value is negative 14.