Question

Consider the polynomial function f defined by
f(x) = x^3 + 2x − 8
Let the inverse of f be g.Find the best linear approximation to g when the argument
of g is close to -5.

Answer 1

I've calculated that derivative of g at -5 would be 1/5. What do I do next?

Answer 2

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

Answer 3

An example problem to your question. I hope
that will address your question, emcrazy14\(\small : ) \)
\(\color{#0000ff }{\huge \displaystyle_{_{_\text{_________________________________}}} }\)
\(\color{#0000ff }{ \displaystyle \bf Problem: }\)

Consider the (polynomial) function \(\color{#000000 }{ \displaystyle f}\) defined by\({\tiny\\[0.5em]}\)
\(\color{#000000 }{ \displaystyle f(x)=x^3-4x^2+10 }\), and let \(\color{#000000 }{ \displaystyle f^{-1}(x)=g(x) }\)\({\tiny\\[0.5em]}\)
Find the best linear approximation to \(\color{#000000 }{ \displaystyle g}\), near\({\tiny\\[0.5em]}\)
the point \(\color{#000000 }{ \displaystyle (3.45,~3.45) }\).
\(\color{#0000ff }{\huge \displaystyle_{_{_\text{_________________________________}}} }\)
\(\color{#0000ff }{ \displaystyle \bf Solution: }\)

\(\color{red}{_■}\) Step 1\( : \) Find the inverse of \(\color{#000000 }{ \displaystyle f }\).

\(\color{#000000 }{ \displaystyle f(x)=x^3-4x^2+10 }\)
I am putting \(\color{#000000 }{ \displaystyle y }\) instead of \(\color{#000000 }{ \displaystyle f(x) }\) for convenience.
\(\color{#000000 }{ \displaystyle y=x^3-4x^2+10 }\)
Then, I swap the x and y (with each other).
\(\color{#000000 }{ \displaystyle x=y^3-4y^2+10 }\)

Solve this for \(\color{#000000 }{ \displaystyle y }\) is painful, be have already
found the inverse technically. So we just have
to remember that really \(\color{#000000 }{ \displaystyle f^{-1}(x)=g(x)=y }\)
and we need to approximate this linearly as
at \(\color{#000000 }{ \displaystyle x=3.45 }\)
So we will leave it as, \(\color{#000000 }{ \displaystyle x=y^3-4y^2+10 }\)
and use implicit differentiation instead.


\(\color{red}{_■}\) Step 2\( : \) Differentiate \(\color{#000000 }{ \displaystyle x=y^3-4y^2+10 }\).

CHAIN RULE NOTE\( : \)
Note, that the chain rule of "y-prime" is applied.
Just like the derivative of \((x^2+3)^{100}\) is NOT just
\(100(x^2+3)^{99}\), RATHER, you differentiate the part
inside as well and multiply (via Chain Rule) and so
you would get \(100(x^3+3)^{99}\color{blue}{\times 3x^2 }\). The blue comes
from the "Chain Rule".
That simplifies to \(300x^2(x^3+3)^{99}\), but that is NOT
the point. THE POINT IS: when you differentiate \(y\), you
also multiply times the derivative of y (times y'), since
y is also a function of x.
Therefore, when you differentiate (for instance) \(y^4\)
(with respect to x), you get, \(4y^3\color{blue}{\times y'}\).

DIFFERENTIATING the function \(\color{#000000 }{ \displaystyle x=y^3-4y^2+10 }\) \( : \)
\(\color{#000000 }{ \displaystyle 1=3y^2y'-8yy'+0 }\)
Then, simplifying and solving for y',
\(\color{#000000 }{ \displaystyle 1=3y^2y'-8yy' }\)
\(\color{#000000 }{ \displaystyle 1=(3y^2-8y)y' }\)
\(\color{#000000 }{ \displaystyle y'=1/(3y^2-8y) }\)
Just in case y' notation confuses you, that is also
an equivalent to saying
\(\color{#000000 }{ \displaystyle \frac{dy}{dx}=\frac{1}{3y^2-8y} }\)

\(\color{red}{_■}\) Step 3\( : \) Find slope at \((3.45, 3.45)\).
To find the slope of the function at the point
(3.45, 3.45), We plug in x=3.45 and y=3.45
(There isn't an "x" in the derivative, so we just plug
y=3.45)

\(\color{#000000 }{ \displaystyle g'(3.45,3.45)=\frac{1}{3(3.45)^2-8(3.45)} =\frac{1}{8.1075}\approx 0.123 }\)
And that is our slope.

\(\color{red}{_■}\) Step 4\( : \) Linearization of \(g\) at \(\color{#000000 }{ \displaystyle (3.45,~3.45) }\).
Given the point (3.45, 3.45)\((3.45, 3.45)\) and slope \(m=0.123\);
y-3.45=0.123(x-3.45)\(y-3.45=0.123(x-3.45)\)
CHECKING:
(In the graph I entered the EXACT slope)
https://www.desmos.com/calculator/3sdq7ik3df