Let g denote the inverse of the function find f(1) and find g'(f(1))

A

akarbarz

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Hello, I am stuck on problem, the problem is as follows.
Let g denote the inverse of the function f(x) = x^3 + x + 2

Part A: find f(1)
which is easy, f(1) = 4.

Part B: find g'(f(1))

This is tricky I first tried to take the inverse of the function, but that is a complete mess. A friend suggested
I take the derivative of the function and then take the reciprocal of the derivative and plug in 4 to find my answer. Would this be correct?
Thanks in advance!
 
akarbarz said:
Hello, I am stuck on problem, the problem is as follows.
Let g denote the inverse of the function f(x) = x^3 + x + 2

Part A: find f(1)
which is easy, f(1) = 4.

Part B: find g'(f(1))

This is tricky I first tried to take the inverse of the function, but that is a complete mess. A friend suggested I take the derivative of the function and then take the reciprocal of the derivative and plug in 4 to find my answer. Would this be correct?
Click to expand...
Didn't your textbook and instructor cover this? Didn't they provide you with a formula similar to "Fact 2" about 1/3 of the way down this page? Just plug into that formula. The answer is one step.

If you get stuck, please reply showing your work, so we can see where you're getting bogged down. Thank you! ;)
 
stapel said:
Didn't your textbook and instructor cover this? Didn't they provide you with a formula similar to "Fact 2" about 1/3 of the way down this page? Just plug into that formula. The answer is one step.

If you get stuck, please reply showing your work, so we can see where you're getting bogged down. Thank you! ;)
Click to expand...


Thanks!
I couldn't seem to find the formula in my book plus (I only slightly remembered there was one for it but was not sure)
Thanks to formula heres what I got,
g'(f(1)) = 1/(3x^2+1)
= 1/ [(3)(4)^2+1]
= 1/49
Would this be correct?
Thanks again.
 
akarbarz said:
Thanks!
I couldn't seem to find the formula in my book plus (I only slightly remembered there was one for it but was not sure)
Thanks to formula heres what I got,
g'(f(1)) = 1/(3x^2+1)
= 1/ [(3)(4)^2+1]
= 1/49
Would this be correct?
Thanks again.
Click to expand...

EDIT:
I figured out I was incorrect and the correct answer is 1/4
Thanks again!
 
akarbarz said:
I couldn't seem to find the formula in my book plus (I only slightly remembered there was one for it but was not sure)
Thanks to formula heres what I got,
g'(f(1)) = 1/(3x^2+1)
= 1/ [(3)(4)^2+1]
= 1/49
Would this be correct?
Click to expand...
The formula given in the page at the link may be restated, for the purposes of this exercise, as:


If g(x) and f (x) are inverses of each other, then the following holds:

. . . . .\(\displaystyle f'(x)\, =\, \dfrac{1}{g'(f(x))}\)


Then, given the original function statement:

. . . . .\(\displaystyle f(x)\, =\, x^3\, +\, x\, +\, 2\)

...we get a derivative of:

. . . . .\(\displaystyle f'(x)\, =\, 3x^2\, +\, 1\)

Since:

. . . . .\(\displaystyle f'(x)\, =\, \dfrac{1}{g'(f(x))}\)

...then:

. . . . .\(\displaystyle g'(f(x))\, =\, \dfrac{1}{f'(x)}\)

Evaluating at x = 1, we get:

. . . . .\(\displaystyle f'(1)\, =\, 3(1)^2\, +\, 1\, =\, 3\, +\, 1\, =\, 4\)

Then:

. . . . .\(\displaystyle g'(f(1))\, =\, \dfrac{1}{f'(1)}\, =\, ...\)


I do not see how you got your value...? Did you conflate the value of f (x) with the value of x itself?
 
stapel said:
The formula given in the page at the link may be restated, for the purposes of this exercise, as:


If g(x) and f (x) are inverses of each other, then the following holds:

. . . . .\(\displaystyle f'(x)\, =\, \dfrac{1}{g'(f(x))}\)


Then, given the original function statement:

. . . . .\(\displaystyle f(x)\, =\, x^3\, +\, x\, +\, 2\)

...we get a derivative of:

. . . . .\(\displaystyle f'(x)\, =\, 3x^2\, +\, 1\)

Since:

. . . . .\(\displaystyle f'(x)\, =\, \dfrac{1}{g'(f(x))}\)

...then:

. . . . .\(\displaystyle g'(f(x))\, =\, \dfrac{1}{f'(x)}\)

Evaluating at x = 1, we get:

. . . . .\(\displaystyle f'(1)\, =\, 3(1)^2\, +\, 1\, =\, 3\, +\, 1\, =\, 4\)

Then:

. . . . .\(\displaystyle g'(f(1))\, =\, \dfrac{1}{f'(1)}\, =\, ...\)


I do not see how you got your value...? Did you conflate the value of f (x) with the value of x itself?
Click to expand...

I got my value by finding f(1) which equals four and then using 4 in that formula, when I should have been using 1. I ended up with a new answer of 1/4. Thanks for your help!
 
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