In Exercises 67–72, you will explore some functions and their ...

Question

In Exercises 67–72, you will explore some functions and their inverses together with their derivatives and tangent line approximations at specified points. Perform the following steps using your CAS:

d. Find the equation for the tangent line to g at the point (f(x_0), x_0) located symmetrically across the 45° line y=x (which is the graph of the identity function). Use Theorem 1 to find the slope of this tangent line.

67. y= √(3x-2), 2/3 ≤ x ≤ 4, x_0=3

Accepted Answer

Hello there. Today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Given that G is the inverse function of F. Find the equation of the tangent line to the graph of G at the point P is equal to parentheses of F X 0, X0. F of X is equal to the square root of 2 x + 54 x is greater than or equal to -5 divided by 2. X subscript 0 is equal to 2. OK, so it appears for this particular problem, we're ultimately asked to determine what the equation of the tangent line is to the graph of this function of G, this graph of G, at the point that the specific point that is given to us by the prom itself, provided we are given a specific expression for the function F of X. And we're told that X is greater than or equal to -5 divided by 2, and we're told that X subscript 0 is equal to 2. So based on all this information that is provided to us, we're ultimately trying to determine what the equation of the tangent line is. So, now that we know what we're ultimately trying to solve for, our first step that we need to take is we need to let. F of X, this function of FX be equal to the square root of 2 x + 5, and we need to note that X subscript 0, which I'll call X0 for short, X0 will be equal to 2. Our second step now is we need to compute F of 2, so we're plugging in a value of 2 in for x. So F of 2 is equal to the square root of 2 multiplied by 2 + 5. Which will be equal to, so 2 multiplied by 2 is 4, so 4 plus 5 is 9, so the square root of 9 is equal to simply 3. So our 3rd step now is we need to recall a note. The point is on F. So the point on F will be. Parentheses 2.3, and its reflection will be across y is equal to x will be. Parentheses 3,2. So once again, the point on F is 2,3, and its reflection across Y is equal to X is 3,2. And for our fourth step, we need to solve for Y is equal to the square root of 2 X + 5 or X. So when we do that, we will find that X is equal to, so once again, we're rearranging this expression to isolate and solve for X, and when we do, we will find that X is equal to Y2 minus 5 divided by 2. So that will mean that G of X, this function for GX will be equal to X2 minus 5 divided by 2. Fantastic. So, moving right along here, step 5 is we need to recall and use the formula that states that F to the power of -1 Y0 will be equal to 1 divided by FX0. Where we need to know where Y, so once again, where Y0 is equal to F of X0. Fantastic. So, moving right along here. So, our sixth steps, this will be step number 6, is we need to compute F where F X denotes the first derivative of FX. So F X will be equal to 1/2, 1 divided by 2, multiplied by parentheses of 2 X plus 5. To the power of -1/2. Multiplied by 2. So once again, multiplied by 2, and this will be equal to 1 divided by the square root of 2 multiplied by X or 2 X plus 5. Fantastic. So, step number 7 is we need to evaluate F2. So once again, we're plugging in a value of 2 in for X, and F2 is equal to 1 divided by the square root of 9, which will be equal to 1 divided by 3. Step number 8 is we need to note that the slope of the tangent line 2G, so once again the slope which. Let's make a note The slope. So now we're focusing on the slope. So, the slope of the tangent line 2G at parentheses 3,2 is 1 divided by F2, which will be equal to 3. So, step number 9, we need to note that the equation of the tangent line. At parentheses 3.2 is Y minus 2 is equal to 3 multiplied by parentheses of x minus 3, and when we rearrange this expression to island sulfur Y by itself, we will find that Y is equal to 3 x minus 7, and this is our final answer. Hooray, we did it. So going back to look at the prom itself to recap what we've learned, our final answer for the tangent line equation, without a doubt is Y is equal to 3X minus 7. And that's it. That is our final answer. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

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