Inverse functions and their derivatives.

**Daithi19** · November 23rd 2011, 01:35 PM

So our lecturer gives us an assignment to be completed every week with ten questions on it. The assignments are unique to each student with the same type of questions but with different numbers for each student, but there is also a sample paper that we can ask tutors for help with.
There are two questions that I'm completely clueless with and need to know how to do, I'm not so much interested in the answer as I am in the technique of how to solve the question so if you could describe the solution step by step I'd appreciate it!

And just so I can't be accused of trying to get people to do my assignment here's a link to assignment page log in as a guest and choose sheet 5 and you'll see both questions there.
http://alberti.nuigalway.ie:8125/exquery.html

**TheEmptySet** · November 23rd 2011, 01:59 PM

Originally Posted by Daithi19

So our lecturer gives us an assignment to be completed every week with ten questions on it. The assignments are unique to each student with the same type of questions but with different numbers for each student, but there is also a sample paper that we can ask tutors for help with.
There are two questions that I'm completely clueless with and need to know how to do, I'm not so much interested in the answer as I am in the technique of how to solve the question so if you could describe the solution step by step I'd appreciate it!

And just so I can't be accused of trying to get people to do my assignment here's a link to assignment page log in as a guest and choose sheet 5 and you'll see both questions there.
http://alberti.nuigalway.ie:8125/exquery.html

For the first one

The function is injective because its derivative is always positive so the function is always increasing.

Now we need to solve the equation

$-3=f(x)=2x^3+11x-3 \iff x(2x^2+11)=0$ the only real solution to this equation is $x=0$

Now we can use the inverse function theorem: That is

$\frac{d}{dx}f^{-1}(x) \bigg|_{x=b}=\frac{1}{f'(a)}$

Where $b=f(a)$

Using the calculation from above we have that

$\frac{d}{dx} f^{-1}(x) \bigg|_{x=-3}=\frac{1}{f'(0)}=\frac{1}{6(0)^2+11}=\frac{1}{11 }$

Now you try the next one.

**Siron** · November 23rd 2011, 02:02 PM

Question 1). First of all, do you know what's an injective function? Then I think it can be useful to use this rule:
$\frac{d f^{-1}}{dx}(f(x))=\frac{1}{f'(x)}$

**skeeter** · November 23rd 2011, 02:13 PM

if $f(x)$ and $g(x)$ are inverse functions, then

$f[g(x)] = x$

$f'[g(x)] \cdot g'(x) = 1$

$g'(x) = \frac{1}{f'[g(x)]}$

note that you can do the same with $g[f(x)] = x$ and arrive at the conclusion that $f'(x) = \frac{1}{g'[f(x)]}$

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