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Given that π of π₯ is equal to eight π₯ minus eight if π₯ is less than negative two and π of π₯ is equal to ππ₯ cubed if π₯ is greater than or equal to negative two is a continuous function, find the value of π.
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What can be said of the differentiability of π at π₯ is equal to negative two?
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In this question, weβre given the piecewise-defined function π of π₯, and weβre told that this is a continuous function.
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We need to use this to find the value of π.
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By looking at the definition of π of π₯, we can see something interesting.
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When π₯ is less than negative two, our function π of π₯ is exactly equal to the linear function eight π₯ minus eight.
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And when π₯ is greater than or equal to negative two, we can see that our function π of π₯ is exactly the same as the cubic function ππ₯ cubed.
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And we know both of these are examples of polynomial functions.
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So theyβre continuous for all real values of π₯.
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In other words, π of π₯ is a piecewise-defined function where each piece is continuous.
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We call these piecewise continuous functions.
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And we know piecewise continuous functions will be continuous everywhere, except maybe at the endpoints of our intervals.
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So in this case, it doesnβt matter what the value of π is.
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We know π will always be continuous for all values of π₯ less than negative two and all values of π₯ greater than negative two.
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We just donβt know what happens when π₯ is equal to negative two.
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Of course, weβre told that π of π₯ is a continuous function.
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So it must be continuous when π₯ is equal to negative two.
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So to help us find the value of π, letβs recall the definition of continuity at π₯ is equal to negative two.
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Thereβs a lot of different ways of writing this.
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Weβll recall the following method.
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We say that π is continuous at π₯ is equal to negative two if the following three conditions are held.
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First, π evaluated at negative two must be defined.
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Another way of saying this is saying that negative two must be in the domain of our function π of π₯.
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Next, we need the limit as π₯ approaches negative two from the left of π of π₯ to be equal to the limit as π₯ approaches negative two from the right of π of π₯.
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Equivalently, youβll often see this written as the limit as π₯ approaches negative two of π of π₯ exists.
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However, the version weβve written down is more useful for piecewise-defined functions.
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Finally, we need the limit as π₯ approaches negative two of π of π₯ equal to π evaluated at negative two.
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And itβs worth pointing out that the limit as π₯ approaches negative two of π of π₯ will be equal to both the limit as π₯ approaches negative two from the left of π of π₯ and the limit as π₯ approaches negative two from the right of π of π₯.
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So we need to check all three of these conditions.
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Letβs start with the first condition.
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We need to show that negative two is in the domain of our function π of π₯.
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And we can just do this by using the piecewise definition of π of π₯.
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π evaluated at negative two is equal to π times negative two all cubed, which we can simplify to give us negative eight π.
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So negative two is in the domain of our function π of π₯.
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To check the second part of our continuity condition, weβll check each of these limits separately.
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Letβs start with the limit as π₯ approaches negative two from the left of π of π₯.
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Since π₯ is approaching negative two from the left, all of our values of π₯ will be less than negative two.
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And from our piecewise definition of the function π of π₯, when π₯ is less than negative two, π of π₯ is exactly equal to the linear function eight π₯ minus eight.
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So because these two functions are exactly equal when π₯ is less than negative two, their limits as π₯ approach negative two from the left will also be equal.
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And of course this is now the limit of a linear function.
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So we can just do this by using direct substitution.
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Substituting in π₯ is equal to negative two, we get eight multiplied by negative two minus eight.
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And if we calculate this, we see itβs equal to negative 24.
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We now want to calculate the limit as π₯ approaches negative two from the right of π of π₯.
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This time, since our values of π₯ are approaching negative two from the right, all of our values of π₯ will be greater than negative two.
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Once again, from the piecewise definition of π of π₯, we can see when our values of π₯ are greater than or equal to negative two, π of π₯ is exactly equal to the cubic ππ₯ cubed.
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So because these two functions are exactly the same when π₯ is greater than or equal to negative two, their limits as π₯ approach negative two from the right will also be equal.
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But now we can see weβre calculating the limit as π₯ approaches negative two from the right of a cubic polynomial.
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So once again, we can evaluate this by direct substitution.
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We get π times negative two all cubed, which of course we can simplify to give us negative eight π.
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Remember, weβre told that π of π₯ is a continuous function.
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So in particular, it must be continuous when π₯ is equal to negative two.
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So by our second continuity condition, the limit as π₯ approaches negative two from the left of π of π₯ must be equal to the limit as π₯ approaches negative two from the right of π of π₯.
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In other words, we must have negative 24 is equal to negative eight π.
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And if we solve this equation by dividing through by negative eight, we see that π must be equal to three.
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And setting π equal to three means both the limit as π₯ approaches two from the left of π of π₯ and the limit as π₯ approaches two from the right of π of π₯ are both equal to negative 24.
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In fact, setting π equal to three also sets π evaluated at negative two equal to negative 24, which we see is also our third continuity condition.
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Therefore, if we set π equal to three, our function π of π₯ will be continuous for all real values of π₯.
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The next part of this question wants us to discuss the differentiability of our function π when π₯ is equal to negative two.
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To do this, letβs clear some space.
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Recall that the derivative of π at π₯ is equal to negative two is supposed to tell us the slope of the tangent line when π₯ is equal to negative two.
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And we can see that π₯ is equal to negative two is the endpoint of our interval.
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In other words, itβs the point where the two pieces of our piecewise-defined function π of π₯ join up.
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Now, weβve already shown that these two pieces join at the same point.
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We showed this when we were checking the continuity of π at π₯ is equal to negative two.
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However, for our tangent line to have a well-defined slope at this point, weβre going to need the slope from the left and the slope from the right to be equal.
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Now, we could do this from first principles straight from the definition of derivatives.
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However, thereβs a simpler method because weβre given a piecewise continuous function and we know how to differentiate each piece of this function.
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We can actually find an expression for π prime of π₯ by first differentiating each piece.
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We get that π prime of π₯ is equal to the derivative of eight π₯ minus eight with respect to π₯ if π₯ is less than negative two and π prime of π₯ is equal to the derivative of three π₯ cubed with respect to π₯ if π₯ is greater than negative two.
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And itβs worth reiterating at this point.
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This wonβt tell us the differentiability of π of π₯ at the endpoints of our intervals.
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So in our case, this hasnβt yet told us whether π of π₯ is differentiable when π₯ is equal to negative two.
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However, it will help us to check whether the slope matches up from the left and from the right.
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We can now evaluate both of these derivatives.
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First, eight π₯ minus eight is a linear function.
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So its derivative with respect to π₯ will just be equal to the coefficient of π₯, which in this case is eight.
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Next, to differentiate three π₯ cubed with respect to π₯, weβll use the power rule for differentiation.
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We need to multiply by the exponent of π₯ and then reduce this exponent by one.
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This gives us nine π₯ squared.
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So weβve now found the slope of π of π₯ when π₯ is less than negative two and the slope of π of π₯ when π₯ is greater than negative two.
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We can use this to find the slope of π of π₯ as π₯ approaches negative two from the left and the slope of π of π₯ as π₯ approaches negative two from the right.
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Letβs start with the slope of π of π₯ as π₯ approaches negative two from the left.
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When π₯ is less than negative two, we can see the slope of π of π₯ is the constant eight.
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So this value is equal to eight.
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We can do the same to find the slope of π of π₯ as π₯ approaches negative two from the right.
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Since π₯ is approaching negative two from the right, our values of π₯ are greater than negative two.
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And we know in this case the slope of π of π₯ will be nine π₯ squared.
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And of course this is a quadratic polynomial.
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So we can do this by using direct substitution.
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We get nine times negative two all squared.
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And if we evaluate this expression, we get 36.
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Therefore, what weβve shown is the slope as π₯ approaches negative two from the left of π of π₯ is not equal to the slope as π₯ approaches negative two from the right of π of π₯.
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This tells us that π canβt possibly be differentiable when π₯ is equal to negative two.
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It is also worth pointing out that if these two values were equal, we would also need to check that π was continuous at negative two.
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But we already did this.
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Therefore, given the function π of π₯ is equal to eight π₯ minus eight if π₯ is less than negative two and π of π₯ is equal to ππ₯ cubed if π₯ is greater than or equal to negative two is a continuous function, we were able to show the value of π must be equal to three.
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We were also able to show that our function π of π₯ was not continuous when π₯ was equal to negative two.