Video: APCALC03AB-P1B-Q32-308147084861

Let 𝑓(π‘₯) = π‘₯Β³ + 27. And let 𝑔 be the inverse function of 𝑓. What is the value of 𝑔′(0)?

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Video Transcript

Let 𝑓 of π‘₯ equal π‘₯ cubed plus 27. And let 𝑔 be the inverse function of 𝑓. What is the value of 𝑔 prime of zero?

Now, it would actually be possible to answer this question by first rearranging the equation 𝑓 of π‘₯ to find the inverse function 𝑔. However, this isn’t really the intention of the question. And it may not always be possible to do that. Instead, we need to use what we know about the derivatives of inverse functions. We should recall that if 𝑓 of π‘₯ and 𝑔 of π‘₯ are inverse functions, then 𝑔 prime of π‘₯ is equal to one over 𝑓 prime of 𝑔 of π‘₯. This is actually relatively straightforward to prove. So we’ll just have a quick look at this. If 𝑓 of π‘₯ and 𝑔 of π‘₯ are the inverse functions of one another, then if we compose them in either order. That is, if we take an π‘₯-value, apply 𝑔 of π‘₯, and then apply 𝑓 of π‘₯ to the result. We will get back to our starting π‘₯-value. So we have that 𝑓 of 𝑔 of π‘₯ is equal to π‘₯.

Let’s now see what happens when we take the derivative of each side of this equation with respect to π‘₯. On the right-hand side, the derivative of π‘₯ with respect to π‘₯ is just one. And in order to find the derivative of a composite function, we need to recall the chain rule. The chain rule tells us that the derivative of the composite function 𝑓 of 𝑔 of π‘₯ is 𝑔 prime of π‘₯ multiplied by 𝑓 prime of 𝑔 of π‘₯. We take the derivative of the inner function multiplied by the derivative of the outer function, with the inner function still inside. It’s a straightforward rearrangement of this equation then to see that 𝑔 prime of π‘₯ is equal to one over 𝑓 prime of 𝑔 of π‘₯. And so we’ve proven the given result.

Now we’re not looking for the general function 𝑔 prime of π‘₯. We’re looking to evaluate this when π‘₯ is equal to zero. So substituting π‘₯ equals zero into our general result gives that 𝑔 prime of zero is equal to one over 𝑓 prime of 𝑔 of zero. Let’s consider how we can find 𝑔 of zero without explicitly finding the inverse function 𝑔. As 𝑓 and 𝑔 are the inverse functions of one another, 𝑓 of 𝑔 of zero will be equal to zero. And so 𝑔 of zero is the value in the domain of 𝑓 of π‘₯ that makes 𝑓 of π‘₯ equal to zero. We can therefore solve the equation 𝑓 of π‘₯ equals zero to find the value of 𝑔 of zero. The function 𝑓 of π‘₯ is π‘₯ cubed plus 27. So we have π‘₯ cubed plus 27 equals zero. We subtract 27 from each side and then take the cube root of each side of the equation. Giving π‘₯ equals the cubed root of negative 27, which is equal to negative three. So we know that the value 𝑔 of zero is equal to negative three. Because 𝑓 of negative three, 𝑓 of 𝑔 of zero, is equal to zero.

Next, we need to find the general expression for the first derivative of our function 𝑓 of π‘₯, 𝑓 prime of π‘₯. And we can do this using the power rule of differentiation. The derivative of π‘₯ cubed with respect to π‘₯ is three π‘₯ squared. And the derivative of any constant with respect to π‘₯ is zero. So we have that 𝑓 prime of π‘₯ is equal to three π‘₯ squared. Finally then, we can substitute into our expression for 𝑔 prime of zero. It’s equal to one over 𝑓 prime of 𝑔 of zero. 𝑔 of zero is negative three. And 𝑓 prime of π‘₯ is three π‘₯ squared. So in the denominator, we have three multiplied by negative three squared. That’s one over three times nine, which is equal to one over 27. So by recalling the relationship that exists between the derivatives of inverse functions. We were able to evaluate the derivative of the inverse function 𝑔 at a given π‘₯-value, without explicitly finding the function 𝑔 itself.

We’ve found that the value of 𝑔 prime of zero is one over 27.

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