Video: Find the Equation of a Tangent Line to an Inverse Function- Given Only the Original Function and a Point on the Curve

Consider the function 𝑓(π‘₯) = π‘₯Β³ + 2π‘₯ βˆ’ 1 and 𝑃(11, 2). Find the slope of the tangent line to its inverse function 𝑓⁻¹ at the indicated point 𝑃. Find the equation of the tangent line to the graph of 𝑓⁻¹ at the indicated point 𝑃.

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Video Transcript

Consider the function 𝑓 of π‘₯ is equal to π‘₯ cubed plus two π‘₯ minus one and the point 𝑃 11, two. Find the slope of the tangent line to its inverse function 𝑓 inverse at the indicated point 𝑃. Find the equation of the tangent line to the graph of 𝑓 inverse at the indicated point 𝑃.

The question gives us a function 𝑓 of π‘₯ which is a cubic polynomial. It also gives us a point 𝑃 which is 11, two. The first part of this question wants us to find the slope of the tangent line to the inverse function of 𝑓 at the point 𝑃. First, the slope of a tangent line to a function at a point is just the slope of that function at that point. So the slope we want to find is the slope of our inverse function of 𝑓 at the point 𝑃. And in fact, since we’re looking for the slope of the tangent line to our inverse function at the point 𝑃, the point 𝑃 must lie on the graph of the inverse function of 𝑓. And 𝑃 is the point 11, two.

And since this point lies on the curve 𝑦 is equal to the inverse function of 𝑓 of π‘₯, when we input 11 into our inverse function, we must get two. And of course, since we want to find the slope of our inverse function at 𝑃, we want to find the slope of our inverse function when π‘₯ is equal to 11. We’re now ready to try and find our slope. We know a formula for finding the slope of an inverse function. We know for a function 𝑓, the derivative of the inverse function of 𝑓 at a point π‘Ž is equal to one divided by 𝑓 prime evaluated at the inverse function of 𝑓 of π‘Ž. Provided π‘Ž is in the domain of our inverse function of 𝑓 and 𝑓 prime is not equal to zero at this point.

We want to find the slope of our function 𝑓 when π‘₯ is equal to 11. So we’ll use π‘Ž is equal to 11. This gives us the derivative of the inverse function of 𝑓 at 11 is equal to one divided by 𝑓 prime evaluated at the inverse function of 𝑓 of 11. And we already know the inverse function of 𝑓 of 11; it’s equal to two. So the slope of our tangent line is just one divided by 𝑓 prime of two. So let’s find an expression for 𝑓 prime of π‘₯.

Remember, 𝑓 of π‘₯ is the cubic π‘₯ cubed plus two π‘₯ minus one. So we can differentiate this term by term by using the power rule for differentiation. And this gives us that 𝑓 prime of π‘₯ is equal to three π‘₯ squared plus two. Substituting π‘₯ is equal to two, we get 𝑓 prime of two is equal to three times two squared plus two, which is equal to 14. Substituting this back into our expression, we get that the slope of our inverse function at the point 𝑃 is one divided by 14. And we know that this must also be equal to the slope of our tangent line to the inverse function of 𝑓 at the point 𝑃.

The second part of our question wants us to find the equation of the tangent line to the graph of our inverse function of 𝑓 at the point 𝑃. To start, we’ll recall the general equation of a straight line. A line with slope π‘š, which passes through the point π‘₯ one, 𝑦 one, will have an equation 𝑦 minus 𝑦 one is equal to π‘š times π‘₯ minus π‘₯ one. And in the first part of our question, we already found that the slope of our tangent line is one divided by 14. So we’ll set π‘š to be one divided by 14.

And remember, we’re finding the equation of the tangent line to our graph of 𝑓 inverse at the point 𝑃. So our tangent line must pass through the point 𝑃, so we can just set π‘₯ one to be 11 and 𝑦 one to be two. So we’ll substitute these values for π‘₯ one, 𝑦 one, and π‘š into our general equation of a line. We get 𝑦 minus two is equal to one divided by 14 times π‘₯ minus 11. Next, we’ll distribute one fourteenth over our parentheses and then add two to both sides of this equation. We get 𝑦 is equal to one over 14 times π‘₯ minus 11 over 14 plus two. And we can then simplify this to get 𝑦 is equal to one over 14 times π‘₯ minus 17 over 14.

Therefore, we’ve shown for the function 𝑓 of π‘₯ is equal to π‘₯ cubed plus two π‘₯ minus one and the point 𝑃 is equal to 11, two, the tangent line to the inverse function of 𝑓 at the point 𝑃 will have a slope of one divided by 14. And the equation of this tangent line will be 𝑦 is equal to one over 14π‘₯ minus 17 divided by 14.

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