Inverse Function from Clep Samplr

[smw0130]

This is the problem from the Calculus Prep:

41. Let \(\displaystyle f(x)\, =\, x^3\, +\, x.\) If \(\displaystyle h\) is the inverse function of \(\displaystyle f,\) then \(\displaystyle h'(2)\, =\)

(A) \(\displaystyle \dfrac{1}{13}\) . . . (B) \(\displaystyle \dfrac{1}{4}\) . . . (C) \(\displaystyle 1\) . . . (D) \(\displaystyle 4\) . . . (E) \(\displaystyle 13\)

But I dont Understand how to get the answer 1/4.

I also have looked up answers on the other websites like http://www.mathgoodies.com/forums/topic.asp?ARCHIVE=true&TOPIC_ID=32883ID=32883&h=lAQEV_mfm andhttp://math.stackexchange.com/questions/60907/inverse-of-y-x3-x. Can someone please help me figure out how to answer this problem.

-Sean

 

[stapel]

41. Let \(\displaystyle f(x)\, =\, x^3\, +\, x.\) If \(\displaystyle h\) is the inverse function of \(\displaystyle f,\) then \(\displaystyle h'(2)\, =\)

(A) \(\displaystyle \dfrac{1}{13}\) . . . (B) \(\displaystyle \dfrac{1}{4}\) . . . (C) \(\displaystyle 1\) . . . (D) \(\displaystyle 4\) . . . (E) \(\displaystyle 13\)

But I dont Understand how to get the answer 1/4.

What formula does your book give you for the relationship(s) between a function, its inverse, its derivative, and the derivative of its inverse?

 

[smw0130]

The inverse of a function is its reciprocal and the derivative of a function is its slope. So the derivative of the inverse must be the reciprocal of its slope.

 

[pka]

This is the problem from the Calculus Prep:
41. Let \(\displaystyle f(x)\, =\, x^3\, +\, x.\) If \(\displaystyle h\) is the inverse function of \(\displaystyle f,\) then \(\displaystyle h'(2)\, =\)
(A) \(\displaystyle \dfrac{1}{13}\) . . . (B) \(\displaystyle \dfrac{1}{4}\) . . . (C) \(\displaystyle 1\) . . . (D) \(\displaystyle 4\) . . . (E) \(\displaystyle 13\)
But I dont Understand how to get the answer 1/4.

By definition, \(\displaystyle h\circ f(x) =x\) so \(\displaystyle [h\circ f(x)]' =1\) or \(\displaystyle (h'\circ f(x))=\dfrac{1}{f'(x)} \).

You know that \(\displaystyle f(1) =2\) so \(\displaystyle (h'(2))=\dfrac{1}{f'(1)}=~? \)

 

[HallsofIvy]

The inverse of a function is its reciprocal and the derivative of a function is its slope. So the derivative of the inverse must be the reciprocal of its slope.

No, the first part of this is not true. The inverse of a function is NOT its "reciprocal!

For example, if f(x)= 3x- 2, its "reciprocal" is \(\displaystyle \frac{1}{f(x)}= \frac{1}{3x- 2}\) while its "inverse" is \(\displaystyle f^{-1}(x)= \frac{x+ 2}{3}\). That is \(\displaystyle f(x)\left(\frac{1}{f(x)}\right)= 1\)- the "reciprocal" is the "multiplicative inverse"- while \(\displaystyle f(f^{-1}(x))= f^{-1}(f(x))= x\)- it is the "composition inverse". Composition being more important than multiplication for functions, we use just the word "inverse" for composition.

Notice that the derivatve of \(\displaystyle 3x- 2\) is "3" and the derivative of \(\displaystyle \frac{x+ 2}{3}\) is 1/3 so the derivative of the inverse function is indeed the reciprocal of the derivative of the original function. But the derivative of \(\displaystyle \frac{1}{3x- 2}\) is \(\displaystyle \frac{-3}{(3x- 2)^2}\).