Video: Evaluating Inverse Function of a Cubic Function

For the function 𝑓(π‘₯) = 2π‘₯Β³ + 6π‘₯ + 10 at π‘Ž = 2, find (𝑓⁻¹)β€²(π‘Ž).

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Video Transcript

For the function 𝑓 of π‘₯ is equal to two π‘₯ cubed plus six π‘₯ plus 10 as π‘Ž is equal to two, find the inverse function of 𝑓 prime evaluated at π‘Ž.

The question gives us a function 𝑓 of π‘₯, which is a cubic. And it wants us to find the first derivative of our inverse function of 𝑓 evaluated at π‘Ž, where π‘Ž is equal to two. And we have a formula for finding the derivative of the inverse function of 𝑓 at π‘₯ is equal to π‘Ž. This is equal to one divided by 𝑓 prime of π‘₯ evaluated at the inverse function of 𝑓 of π‘Ž. And this is valid if π‘Ž is in the domain of our inverse function of 𝑓 and 𝑓 prime is not equal to zero at this value. From our question, we can see we’re asked to find the derivative of the inverse function of 𝑓 evaluated at π‘Ž is equal to two. So we’ll use π‘Ž is equal to two in our formula.

We can now see there are two things we need to use this formula. First, we need an expression for 𝑓 prime of π‘₯. We also need to know what the inverse function of 𝑓 evaluated at two is equal to. Let’s start by finding an expression for 𝑓 prime of π‘₯. Since 𝑓 of π‘₯ is a cubic, we can do this term by term by using the power rule for differentiation. Applying this, we get that 𝑓 prime of π‘₯ is equal to six π‘₯ squared plus six. Now, we need to find an expression for our inverse function of 𝑓 evaluated at two. Let’s call this value 𝑏. One way of finding an expression for our value of 𝑏 is to apply 𝑓 to both sides of this equation. We get that 𝑓 evaluated at the inverse function of 𝑓 evaluated at two is equal to 𝑓 evaluated at 𝑏.

But remember, our function 𝑓 and our function 𝑓 inverse are inverse functions. So 𝑓 compose with 𝑓 inverse is the identity function. It doesn’t actually change our value. So it’s equal to two. And remember, we’re told that 𝑓 evaluated at π‘₯ is equal to two π‘₯ cubed plus six π‘₯ plus 10. So to find 𝑓 evaluated at 𝑏, we just substitute π‘₯ is equal to 𝑏 into our expression for 𝑓 of π‘₯. We get two 𝑏 cubed plus six 𝑏 plus 10. And now, we can see this is just a cubic equation for 𝑏. To solve this for 𝑏, we’ll start by subtracting two from both sides of our equation. This gives us zero is equal to two 𝑏 cubed plus six 𝑏 plus eight.

Next, we’ll divide both sides of our equation through by two. This gives us zero is equal to 𝑏 cubed plus three 𝑏 plus four. Now, by inspection, we can find that if we substitute 𝑏 is equal to negative one, we get negative one cubed plus three times negative one plus four, which is equal to zero. In other words, 𝑏 is equal to negative one is a root of our cubic. So by the factor theorem, 𝑏 plus one must be a factor of this cubic. So we’ll take a factor of 𝑏 plus one outside of our cubic. We could do this by using algebraic division. However, we also know a cubic factors to give a linear multiplied by a quadratic. And we know that 𝑏 multiplied by our term for 𝑏 squared must be equal to 𝑏 cubed. So this must be equal to 𝑏 squared.

Next, we know if we multiply our two constant terms together, we must get four. Finally, our quadratic could have a term for 𝑏. We’ll call this 𝑐 times 𝑏. And to find the value of 𝑐, we’ll equate coefficients of 𝑏 squared. In our cubic, we have no coefficient of 𝑏 squared. So we have zero 𝑏 squared is equal to 𝑏 squared plus 𝑐𝑏 squared. And we can solve this equation. We get 𝑐 is equal to negative one. So we have our cubic is equal to 𝑏 plus one times 𝑏 squared minus 𝑏 plus four. And if we try to factor this quadratic using any method, we’ll notice there are no solutions.

For example, checking the discriminant in the quadratic formula gives us negative one squared minus four times one times four, which we can evaluate to give us negative 15, which tells us that our quadratic has no real roots. So the only possible value for 𝑏 is equal to negative one. So we’ve shown that two is in the domain of our inverse function of 𝑓. And 𝑓 inverse evaluated at two is equal to negative one. We’re now ready to find 𝑓 inverse prime evaluated at two. It’s equal to one divided by 𝑓 prime evaluated at the inverse function of 𝑓 of two. And we showed the inverse function of 𝑓 evaluated at two is equal to negative one.

So writing this in our formula, we get one divided by 𝑓 prime evaluated at negative one. And we found that 𝑓 prime of π‘₯ is equal to six π‘₯ squared plus six. Substituting π‘₯ is equal to negative one into our expression for 𝑓 prime of π‘₯, we get one divided by six times negative one squared plus six. And this simplifies to give us one divided by six plus six, which is equal to one twelfth. Therefore, we’ve shown if 𝑓 of π‘₯ is equal to two π‘₯ cubed plus six π‘₯ plus 10 and π‘Ž is equal to two, then the derivative of the inverse function of 𝑓 evaluated at π‘Ž is equal to one twelfth.

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