Question

derivatives of inverse functions

1. let ( f(x) = x^5 + 2x^3 + x - 1 ).

a. find ( f(1) ) and ( f(1) ).
b. find ( f^{-1}(3) ) and ( (f^{-1})(3) )

1. determine the equation of the tangent line to ( f^{-1}(x) ) at the point where ( x = 3 ) on ( f^{-1} ), given that ( f(2) = 3 ) and ( f(2) = 5 ).
2. determine the equation of the tangent line to ( f^{-1}(x) ) at ( f^{-1}(-6) = 0 ) if ( f(x) = -5 + 2x - cos x ).
3. determine the slope of the tangent line to ( f^{-1}(x) ) at ( f^{-1}(80) = 2 ) if ( f(x) = 3x^5 - 5x^3 + 12x ).
4. multiple choice: let ( f ) be a differentiable function such that ( f(3) = 15 ), ( f(6) = 3 ), ( f(3) = -8 ), and ( f(6) = -2 ). the function ( g ) is differentiable and ( g(x) = f^{-1}(x) ) for all ( x ). what is the value of ( g(3) )?

(a) ( -\frac{1}{2} ) (b) ( -\frac{1}{8} ) (c) ( \frac{1}{6} ) (d) ( \frac{1}{3} )
(e) the value of ( g(3) ) cannot be determined from the information given.

Explanation:

1)

Step1: Compute $f(1)$

Substitute $x=1$ into $f(x)$:
$$f(1)=1^5 + 2(1)^3 + 1 - 1 = 1+2+1-1=3$$

Step2: Compute $f'(x)$ and $f'(1)$

Differentiate $f(x)$:
$$f'(x)=5x^4 + 6x^2 + 1$$
Substitute $x=1$:
$$f'(1)=5(1)^4 + 6(1)^2 + 1=5+6+1=12$$

Step3: Find $f^{-1}(3)$

From Step1, $f(1)=3$, so $f^{-1}(3)=1$.

Step4: Compute $(f^{-1})'(3)$

Use inverse derivative rule: $(f^{-1})'(a)=\frac{1}{f'(f^{-1}(a))}$
$$(f^{-1})'(3)=\frac{1}{f'(f^{-1}(3))}=\frac{1}{f'(1)}=\frac{1}{12}$$

Step1: Identify $f^{-1}(3)$

Given $f(2)=3$, so $f^{-1}(3)=2$.

Step2: Compute slope of tangent

Use inverse derivative rule:
$$(f^{-1})'(3)=\frac{1}{f'(f^{-1}(3))}=\frac{1}{f'(2)}=\frac{1}{5}$$

Step3: Write tangent line equation

Use point-slope form $y-y_1=m(x-x_1)$ with $(x_1,y_1)=(3,2)$:
$$y-2=\frac{1}{5}(x-3)$$
Simplify to slope-intercept form:
$$y=\frac{1}{5}x + \frac{7}{5}$$

Step1: Identify $f(0)$

Given $f^{-1}(-6)=0$, so $f(0)=-6$.

Step2: Compute $f'(x)$ and $f'(0)$

Differentiate $f(x)$:
$$f'(x)=2 + \sin x$$
Substitute $x=0$:
$$f'(0)=2 + \sin 0=2$$

Step3: Find slope of tangent

$$(f^{-1})'(-6)=\frac{1}{f'(f^{-1}(-6))}=\frac{1}{f'(0)}=\frac{1}{2}$$

Step4: Write tangent line equation

Use point-slope form with $(x_1,y_1)=(-6,0)$:
$$y-0=\frac{1}{2}(x+6)$$
Simplify:
$$y=\frac{1}{2}x + 3$$

Answer:

a. $f(1)=3$, $f'(1)=12$
b. $f^{-1}(3)=1$, $(f^{-1})'(3)=\frac{1}{12}$

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