Video Transcript
Consider the function π of π₯ is equal to π₯ to the fifth power minus two π₯ cubed plus three π₯ plus two and the point four, one. Find the slope of the tangent line to its inverse function π inverse at the indicated point π. Find the equation of the tangent line to the graph of the inverse function of π at the indicated point π.
The question gives us a function π of π₯ which is a polynomial of degree five and a point four, one. The first part of this question wants us to find the slope of the tangent line to our inverse function of π at the point π. And if weβre finding the tangent line to our inverse function of π at the point π, then point π must lie on our curve π¦ is equal to the inverse function of π of π₯. And what does this tell us? Well, when we input four into our inverse function of π₯, our output must be one. In other words, the inverse function of π evaluated at four is equal to one.
Now, remember, the slope of our tangent line at this point will just be equal to the slope of our inverse function at this point. And we know a formula for finding the slope of an inverse function. The derivative of our inverse function of π at the point π is equal to one divided by π prime evaluated at the inverse function of π of π. And this assumes that π is in the domain of our inverse function of π and that π prime is not equal to zero at this point.
We want to find the slope of our curve at the point four, one. So weβll set π equal to four. And we already found an expression for the inverse function of π at four. Itβs equal to one. So we get the slope of our inverse function of π at π₯ is equal to four is equal to one divided by π prime of one.
So we need to find an expression for π prime of π₯. And π of π₯ is just a polynomial. So we can do this by using the power rule for differentiation. Applying the power rule for differentiation on each of our terms of π of π₯, we get that π prime of π₯ is equal to five π₯ to the fourth power minus six π₯ squared plus three. So we can now use this expression to evaluate one divided by π prime of one. Itβs equal to one divided by five times one to the fourth power minus six times one squared plus three.
So weβve shown the slope of our inverse function at the point four, one is equal to a half. And remember, this will be the same as the slope of our tangent line.
The second part of our question wants us to find the equation of the tangent line to our graph of the inverse function of π at the point π. The first thing we need to do is remember the general form of the equation of a straight line. A line of slope π which passes through the points π¦ one, π₯ one will have an equation π¦ minus π¦ one is equal to π times π₯ minus π₯ one. And we actually already found the slope of the tangent line to the graph of the inverse function of π at the point π in the first part of our equation. π must be equal to one-half. So we just need to find values for π₯ one and π¦ one.
Remember that the graph of our inverse function of π passes through the point π. And because π is the point four, one, we can just set π₯ one is equal to four and π¦ one equal to one. Substituting these values into the general equation for a straight line, we get π¦ minus one is equal to one-half times π₯ minus four. Distributing one-half over our parentheses, we get π¦ minus one is equal to a half π₯ minus two. Finally, weβll just add one to both sides of this equation. We get π¦ is equal to a half π₯ minus one.
Therefore, weβve shown for the function π of π₯ is equal to π₯ to the fifth power minus two π₯ cubed plus three π₯ plus two and the point π, which is equal to four, one. The slope of the tangent line to the inverse function of π at the point π will be one-half. And the tangent line to the graph of the inverse function of π at the point π will be π¦ is equal to a half π₯ minus one.