What is the value of the inverse derivative at x=f(a)?

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Discussion Overview

The discussion revolves around finding the value of the inverse derivative at a specific point for the function $$f(x)=x^3-6x^2-3$$, particularly at $$x=f(a)$$ where $$a=3$$. Participants explore the implications of the function's behavior, its one-to-one nature, and the necessary steps to isolate the inverse and compute the derivative. The scope includes mathematical reasoning and technical exploration.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant seeks to find $$\frac{df^{-1}}{dx}$$ at $$x=f(a)$$ but struggles with isolating $$y$$ in the inverse function.
  • Another participant suggests verifying that the function is one-to-one on the interval $$[4,\infty)$$ and outlines steps for implicit differentiation and solving for $$x_r$$ where $$f(x)=-3$$.
  • Some participants express uncertainty about the problem's wording, suggesting it should state $$x=f^{-1}(a)$$ and reference the standard formula for the derivative of an inverse function.
  • There is a discussion about the roots of the equation $$x^3-6x^2-3=a$$ for different values of $$a$$, with specific numerical approximations provided.
  • Participants calculate the derivative of the function at the point where $$f^{-1}(a)$$ is found, with differing numerical results presented.
  • One participant offers to share a graph of the function and its inverse, noting the vagueness of the original question and its open interpretation.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the problem and the necessary steps to solve it. There is no consensus on the exact wording of the question or the approach to take, leading to multiple competing views on how to proceed.

Contextual Notes

Some participants note potential typos in the problem statement and the implications of these on the solution process. There are also mentions of the function's one-to-one nature and the need for real roots in the specified domain, which remain unresolved.

Who May Find This Useful

Readers interested in inverse functions, derivatives, and mathematical problem-solving in calculus may find this discussion relevant.

karush
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find the value of $$df^{-1}/dx at $x=f(a)$$
$$f(x)=x^3-6x^2-3$$
$$x \ge 4$$
$$a=3$$

ok the inverse would be
$$x=y^3-6y^2-3$$

but don't see how to isolate $y$
or if we need to
 
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First, I would verify that the given function is one-to-one on $[4,\infty)$. Then I recommend following these steps:

1.) Implicitly differentiation the following:

$$f(x)=x^3-6x^2-3$$

With respect to $f$ and solve for $$\d{x}{f}$$.

2.) Solve (I am assuming $a=-3$):

$$f(x)=-3$$

Call the relevant (must be real and in the stated domain) root $x_r$.

3.) Then we have:

$$\left[f^{-1}\right]'(a)=\left.\d{x}{f}\right|_{x=x_r}$$
 

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I'm thinking there's at least one typo in the problem...I think at the very least it should state:

$$x=f^{-1}(a)$$

Let's look at the standard formula:

$$\left[f^{-1}\right]'(a)=\frac{1}{f'\left(f^{-1}(a)\right)}$$

We know:

$$f\left(f^{-1}(a)\right)=a$$

Let:

$$u=f^{-1}(a)$$

And so we have:

$$f(u)=a$$

Using the given definition of $f$ and the given $a=3$, we have:

$$u^3-6u^2-3=3$$

As you can see, this would be much easier to solve if $a=-3$, but this can be solved. What do you get?
 
MarkFL said:
I'm thinking there's at least one typo in the problem...I think at the very least it should state:

$$x=f^{-1}(a)$$

Let's look at the standard formula:

$$\left[f^{-1}\right]'(a)=\frac{1}{f'\left(f^{-1}(a)\right)}$$

We know:

$$f\left(f^{-1}(a)\right)=a$$

Let:

$$u=f^{-1}(a)$$

And so we have:

$$f(u)=a$$

Using the given definition of $f$ and the given $a=3$, we have:

$$u^3-6u^2-3=3$$

As you can see, this would be much easier to solve if $a=-3$, but this can be solved. What do you get?

what is $f^{-1}(a)$
 
karush said:
what is $f^{-1}(a)$

Suppose we set:

$$f(x)=a$$

This is equivalent to:

$$x=f^{-1}(a)$$

This is what you'll find when solving:

$$x^3-6x^2-3=a$$
 
this?

$\text{so if $a=3$ then}$
$\displaystyle x^3-6x^2-3=3$
$\displaystyle x^3-6x^2=6$
$\text{$x=6.1582$ \, via W|A}$

$\text{or if $a=-3$ then}$
$\displaystyle x^3-6x^2-3=-3$
$\displaystyle x^3-6x^2=0$
$x^2(x-6)=0$
$x=0$
$x=6$
 
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karush said:
this?

$\text{so if $a=3$ then}$
$\displaystyle x^3-6x^2-3=3$
$\displaystyle x^3-6x^2=6$
$\text{$x=6.1582$ \, via W|A}$

Yes, and the exact root is given as:

$$x=2+\sqrt[3]{11+\sqrt{57}}+\sqrt[3]{11-\sqrt{57}}$$

karush said:
$\text{or if $a=-3$ then}$
$\displaystyle x^3-6x^2-3=-3$
$\displaystyle x^3-6x^2=0$
$x^2(x-6)=0$
$x=0$
$x=6$

We would discard the root $x=0$ since we are told $4\le x$. Can you demonstrate the given function is one-to-one on this interval?

At any rate, once we have the value of $f^{-1}(a)$, can you proceed?
 
$\text{I'm not sure what $6.1582$ is supposed to plugged into}$
 
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  • #10
karush said:
$\text{I'm not sure what $6.1582$ is supposed to plugged into}$

You could use this formula:

$$\left[f^{-1}\right]'(a)=\frac{1}{f'\left(f^{-1}(a)\right)}$$

where $a=3,\,f^{-1}(a)\approx6.1582$.
 
  • #11
MarkFL said:
You could use this formula:

$$\left[f^{-1}\right]'(a)=\frac{1}{f'\left(f^{-1}(a)\right)}$$

where $a=3,\,f^{-1}(a)\approx6.1582$.

$\displaystyle \left[f^{-1}\right]'(a)$
$=\frac{1}{f'\left(f^{-1}(a)\right)}$
$=\frac{1}{f'(6.1582)}$
$f'(6.1582)=3(6.11582)^2-12(6.11582)=38.8119$
$1/38.8119 =0.02576$
 
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  • #12
karush said:
$\displaystyle \left[f^{-1}\right]'(a)$
$=\frac{1}{f'\left(f^{-1}(a)\right)}$
$=\frac{1}{f'(6.1582)}$
$f'(6.1582)=3(6.11582)^2-12(6.11582)=38.8119$

I get:

$$f'\left(f^{-1}(a)\right)\approx39.8722$$

W|A - (3(2 + (11 - sqrt(57))^(1/3) + (11 + sqrt(57))^(1/3))^2-12(2 + (11 - sqrt(57))^(1/3) + (11 + sqrt(57))^(1/3)))

And so:

$$\left[f^{-1}\right]'(a)\approx0.02508$$

W|A - 1/(3(2 + (11 - sqrt(57))^(1/3) + (11 + sqrt(57))^(1/3))^2-12(2 + (11 - sqrt(57))^(1/3) + (11 + sqrt(57))^(1/3)))
 
  • #13
I can post a graph of the function, derivative, inverse, and solution in graph forum from MATLAB if that helps anyone.

I must admit, the question is written in a very vague forum. It doesn't define everything how "most mathematicians" would. It leaves a lot of the problem open for interpretation.
 
  • #14
DrWahoo said:
I can post a graph of the function, derivative, inverse, and solution in graph forum from MATLAB if that helps anyone.

I must admit, the question is written in a very vague forum. It doesn't define everything how "most mathematicians" would. It leaves a lot of the problem open for interpretation.

Sure
I did what I could with Desmos but
Could only do so much
 
  • #15
Over 100 views 😎
 
  • #16
karush said:
Sure
I did what I could with Desmos but
Could only do so much

Desmos allows you to plot a function and its inverse...first define the function and its domain:

$$f(x)=x^3-6x^2-3\,\{4\le x\}$$

Then define the inverse:

$$x=f(y)$$

[DESMOS=-37.36896540090182,93.61770297027579,-36.842231263227994,8.599804598412113]f\left(x\right)=x^3-6x^2-3\left\{4\le x\right\};x=f\left(y\right)[/DESMOS]
 
  • #17

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