Answer :
Sure, let's solve each part of the question step-by-step.
### Part a: Finding the Derivative Function [tex]\( f'(x) \)[/tex]
To find the derivative [tex]\( f'(x) \)[/tex] of the function [tex]\( f(x) = \frac{7}{2x+1} \)[/tex], we can use the limit definition of a derivative, but let's use standard differentiation rules for simplicity.
We have [tex]\( f(x) = \frac{7}{2x+1} \)[/tex]. This is a quotient, so we can use the chain rule to differentiate it. Let [tex]\( u = 2x + 1 \)[/tex]. Then:
[tex]\[ f(x) = \frac{7}{u} \][/tex]
To find [tex]\( \frac{d}{dx} \left( \frac{7}{u} \right) \)[/tex], we first find [tex]\( \frac{d}{du} \left( \frac{7}{u} \right) \)[/tex].
[tex]\[ \frac{d}{du} \left( \frac{7}{u} \right) = -\frac{7}{u^2} \][/tex]
Next, we need [tex]\( \frac{du}{dx} \)[/tex]:
[tex]\[ \frac{du}{dx} = 2 \][/tex]
Using the chain rule [tex]\( \frac{d}{dx} = \frac{d}{du} \cdot \frac{du}{dx} \)[/tex]:
[tex]\[ f'(x) = -\frac{7}{u^2} \cdot 2 = -\frac{14}{(2x+1)^2} \][/tex]
So, the derivative function is:
[tex]\[ f'(x) = -\frac{14}{(2x+1)^2} \][/tex]
### Part b: Evaluating [tex]\( f'(a) \)[/tex] for Given Values of [tex]\( a \)[/tex]
Let's now evaluate [tex]\( f'(a) \)[/tex] for [tex]\( a = -\frac{1}{3} \)[/tex] and [tex]\( a = 4 \)[/tex].
1. For [tex]\( a = -\frac{1}{3} \)[/tex]:
[tex]\[ f'\left( -\frac{1}{3} \right) = -\frac{14}{\left( 2 \left( -\frac{1}{3} \right) + 1 \right)^2} \][/tex]
Simplify the denominator:
[tex]\[ 2 \left( -\frac{1}{3} \right) = -\frac{2}{3} \][/tex]
[tex]\[ -\frac{2}{3} + 1 = -\frac{2}{3} + \frac{3}{3} = \frac{1}{3} \][/tex]
So:
[tex]\[ f'\left( -\frac{1}{3} \right) = -\frac{14}{\left( \frac{1}{3} \right)^2} = -\frac{14}{\frac{1}{9}} = -14 \cdot 9 = -126 \][/tex]
Therefore, [tex]\( f' \left( -\frac{1}{3} \right) = -126 \)[/tex].
2. For [tex]\( a = 4 \)[/tex]:
[tex]\[ f'(4) = -\frac{14}{(2 \cdot 4 + 1)^2} \][/tex]
Simplify the expression:
[tex]\[ 2 \cdot 4 + 1 = 8 + 1 = 9 \][/tex]
So:
[tex]\[ f'(4) = -\frac{14}{9^2} = -\frac{14}{81} \][/tex]
Therefore, [tex]\( f'(4) = -\frac{14}{81} \)[/tex].
### Summary
a. The derivative function [tex]\( f'(x) \)[/tex] is:
[tex]\[ f'(x) = -\frac{14}{(2x+1)^2} \][/tex]
b. Evaluating the derivative at the given points:
- [tex]\( f'\left( -\frac{1}{3} \right) = -126 \)[/tex]
- [tex]\( f'(4) = -\frac{14}{81} \)[/tex]
### Part a: Finding the Derivative Function [tex]\( f'(x) \)[/tex]
To find the derivative [tex]\( f'(x) \)[/tex] of the function [tex]\( f(x) = \frac{7}{2x+1} \)[/tex], we can use the limit definition of a derivative, but let's use standard differentiation rules for simplicity.
We have [tex]\( f(x) = \frac{7}{2x+1} \)[/tex]. This is a quotient, so we can use the chain rule to differentiate it. Let [tex]\( u = 2x + 1 \)[/tex]. Then:
[tex]\[ f(x) = \frac{7}{u} \][/tex]
To find [tex]\( \frac{d}{dx} \left( \frac{7}{u} \right) \)[/tex], we first find [tex]\( \frac{d}{du} \left( \frac{7}{u} \right) \)[/tex].
[tex]\[ \frac{d}{du} \left( \frac{7}{u} \right) = -\frac{7}{u^2} \][/tex]
Next, we need [tex]\( \frac{du}{dx} \)[/tex]:
[tex]\[ \frac{du}{dx} = 2 \][/tex]
Using the chain rule [tex]\( \frac{d}{dx} = \frac{d}{du} \cdot \frac{du}{dx} \)[/tex]:
[tex]\[ f'(x) = -\frac{7}{u^2} \cdot 2 = -\frac{14}{(2x+1)^2} \][/tex]
So, the derivative function is:
[tex]\[ f'(x) = -\frac{14}{(2x+1)^2} \][/tex]
### Part b: Evaluating [tex]\( f'(a) \)[/tex] for Given Values of [tex]\( a \)[/tex]
Let's now evaluate [tex]\( f'(a) \)[/tex] for [tex]\( a = -\frac{1}{3} \)[/tex] and [tex]\( a = 4 \)[/tex].
1. For [tex]\( a = -\frac{1}{3} \)[/tex]:
[tex]\[ f'\left( -\frac{1}{3} \right) = -\frac{14}{\left( 2 \left( -\frac{1}{3} \right) + 1 \right)^2} \][/tex]
Simplify the denominator:
[tex]\[ 2 \left( -\frac{1}{3} \right) = -\frac{2}{3} \][/tex]
[tex]\[ -\frac{2}{3} + 1 = -\frac{2}{3} + \frac{3}{3} = \frac{1}{3} \][/tex]
So:
[tex]\[ f'\left( -\frac{1}{3} \right) = -\frac{14}{\left( \frac{1}{3} \right)^2} = -\frac{14}{\frac{1}{9}} = -14 \cdot 9 = -126 \][/tex]
Therefore, [tex]\( f' \left( -\frac{1}{3} \right) = -126 \)[/tex].
2. For [tex]\( a = 4 \)[/tex]:
[tex]\[ f'(4) = -\frac{14}{(2 \cdot 4 + 1)^2} \][/tex]
Simplify the expression:
[tex]\[ 2 \cdot 4 + 1 = 8 + 1 = 9 \][/tex]
So:
[tex]\[ f'(4) = -\frac{14}{9^2} = -\frac{14}{81} \][/tex]
Therefore, [tex]\( f'(4) = -\frac{14}{81} \)[/tex].
### Summary
a. The derivative function [tex]\( f'(x) \)[/tex] is:
[tex]\[ f'(x) = -\frac{14}{(2x+1)^2} \][/tex]
b. Evaluating the derivative at the given points:
- [tex]\( f'\left( -\frac{1}{3} \right) = -126 \)[/tex]
- [tex]\( f'(4) = -\frac{14}{81} \)[/tex]
