Answer :
Sure, let's solve the problem step-by-step.
Given the function:
[tex]$ f(x) = \frac{7}{x^2 - 49} $[/tex]
### (A) Find all critical numbers of [tex]\( f \)[/tex]
To find the critical numbers, we first need to find the derivative of [tex]\( f(x) \)[/tex], which is [tex]\( f'(x) \)[/tex], and then set it equal to zero or find points where it is undefined.
First, rewrite [tex]\( f(x) \)[/tex]:
[tex]$ f(x) = \frac{7}{(x-7)(x+7)} $[/tex]
Let's find the derivative [tex]\( f'(x) \)[/tex]. Using the quotient rule:
If [tex]\( f(x) = \frac{u(x)}{v(x)} \)[/tex], then [tex]\( f'(x) = \frac{u'v - uv'}{v^2} \)[/tex].
Here, [tex]\( u(x) = 7 \)[/tex] and [tex]\( v(x) = x^2 - 49 \)[/tex], thus:
[tex]\[ u' = 0 \][/tex]
[tex]\[ v' = 2x \][/tex]
Applying the quotient rule:
[tex]\[ f'(x) = \frac{0 \cdot (x^2 - 49) - 7 \cdot 2x}{(x^2 - 49)^2} \][/tex]
[tex]\[ f'(x) = \frac{-14x}{(x^2 - 49)^2} \][/tex]
Set [tex]\( f'(x) = 0 \)[/tex]:
[tex]\[ \frac{-14x}{(x^2 - 49)^2} = 0 \][/tex]
This happens when [tex]\( -14x = 0 \)[/tex], which implies:
[tex]\[ x = 0 \][/tex]
We also need to check where [tex]\( f'(x) \)[/tex] is undefined. Since [tex]\( f'(x) = \frac{-14x}{(x^2 - 49)^2} \)[/tex], it will be undefined where the denominator is zero.
[tex]\[ x^2 - 49 = 0 \][/tex]
[tex]\[ x = \pm 7 \][/tex]
So, the function [tex]\( f(x) \)[/tex] is undefined at [tex]\( x = 7 \)[/tex] and [tex]\( x = -7 \)[/tex].
Hence, the only critical number in the domain of [tex]\( f(x) \)[/tex] is [tex]\( x = 0 \)[/tex].
Critical numbers = [tex]\( 0 \)[/tex]
### (B) Use interval notation to indicate where [tex]\( f(x) \)[/tex] is increasing
We use the first derivative [tex]\( f'(x) = \frac{-14x}{(x^2 - 49)^2} \)[/tex] to determine where the function is increasing or decreasing.
Check the sign of [tex]\( f'(x) \)[/tex] on the intervals created by the critical points and points of discontinuity.
1. Interval [tex]\((- \infty, -7)\)[/tex]:
Pick a test point, say [tex]\( x = -8 \)[/tex]:
[tex]\[ f'(-8) = \frac{-14(-8)}{((-8)^2 - 49)^2} > 0 \][/tex]
2. Interval [tex]\((-7, 0)\)[/tex]:
Pick a test point, say [tex]\( x = -1 \)[/tex]:
[tex]\[ f'(-1) = \frac{-14(-1)}{((-1)^2 - 49)^2} > 0 \][/tex]
3. Interval [tex]\((0, 7)\)[/tex]:
Pick a test point, say [tex]\( x = 1 \)[/tex]:
[tex]\[ f'(1) = \frac{-14(1)}{((1)^2 - 49)^2} < 0 \][/tex]
4. Interval [tex]\((7, \infty)\)[/tex]:
Pick a test point, say [tex]\( x = 8 \)[/tex]:
[tex]\[ f'(8) = \frac{-14(8)}{((8)^2 - 49)^2} < 0 \][/tex]
From the test points:
- [tex]\( f(x) \)[/tex] is increasing on the intervals [tex]\((- \infty, -7) \cup (-7, 0)\)[/tex].
Increasing: [tex]\( (- \infty, -7) U (-7, 0) \)[/tex]
### (C) Use interval notation to indicate where [tex]\( f(x) \)[/tex] is decreasing
From the analysis above:
- [tex]\( f(x) \)[/tex] is decreasing on the intervals [tex]\( (0, 7) \cup (7, \infty) \)[/tex].
Decreasing: [tex]\( (0, 7) U (7, \infty) \)[/tex]
### (D) List the [tex]\( x \)[/tex]-coordinates of all local maxima
A local maximum would occur where [tex]\( f'(x) \)[/tex] changes from positive to negative. From the intervals above and the nature of [tex]\( f'(x) \)[/tex]:
There is no point where [tex]\( f'(x) changes from positive to negative. Local maxima: \( NONE \)[/tex]
### (E) List the [tex]\( x \)[/tex]-coordinates of all local minima
A local minimum would occur where [tex]\( f'(x) \)[/tex] changes from negative to positive.
The only critical point is [tex]\( x = 0 \)[/tex], and we observed:
[tex]\( f'(x) \)[/tex] changes from positive to negative, indicating no local minima at [tex]\( x = 0 \)[/tex].
Local minima: [tex]\( NONE \)[/tex]
Given the function:
[tex]$ f(x) = \frac{7}{x^2 - 49} $[/tex]
### (A) Find all critical numbers of [tex]\( f \)[/tex]
To find the critical numbers, we first need to find the derivative of [tex]\( f(x) \)[/tex], which is [tex]\( f'(x) \)[/tex], and then set it equal to zero or find points where it is undefined.
First, rewrite [tex]\( f(x) \)[/tex]:
[tex]$ f(x) = \frac{7}{(x-7)(x+7)} $[/tex]
Let's find the derivative [tex]\( f'(x) \)[/tex]. Using the quotient rule:
If [tex]\( f(x) = \frac{u(x)}{v(x)} \)[/tex], then [tex]\( f'(x) = \frac{u'v - uv'}{v^2} \)[/tex].
Here, [tex]\( u(x) = 7 \)[/tex] and [tex]\( v(x) = x^2 - 49 \)[/tex], thus:
[tex]\[ u' = 0 \][/tex]
[tex]\[ v' = 2x \][/tex]
Applying the quotient rule:
[tex]\[ f'(x) = \frac{0 \cdot (x^2 - 49) - 7 \cdot 2x}{(x^2 - 49)^2} \][/tex]
[tex]\[ f'(x) = \frac{-14x}{(x^2 - 49)^2} \][/tex]
Set [tex]\( f'(x) = 0 \)[/tex]:
[tex]\[ \frac{-14x}{(x^2 - 49)^2} = 0 \][/tex]
This happens when [tex]\( -14x = 0 \)[/tex], which implies:
[tex]\[ x = 0 \][/tex]
We also need to check where [tex]\( f'(x) \)[/tex] is undefined. Since [tex]\( f'(x) = \frac{-14x}{(x^2 - 49)^2} \)[/tex], it will be undefined where the denominator is zero.
[tex]\[ x^2 - 49 = 0 \][/tex]
[tex]\[ x = \pm 7 \][/tex]
So, the function [tex]\( f(x) \)[/tex] is undefined at [tex]\( x = 7 \)[/tex] and [tex]\( x = -7 \)[/tex].
Hence, the only critical number in the domain of [tex]\( f(x) \)[/tex] is [tex]\( x = 0 \)[/tex].
Critical numbers = [tex]\( 0 \)[/tex]
### (B) Use interval notation to indicate where [tex]\( f(x) \)[/tex] is increasing
We use the first derivative [tex]\( f'(x) = \frac{-14x}{(x^2 - 49)^2} \)[/tex] to determine where the function is increasing or decreasing.
Check the sign of [tex]\( f'(x) \)[/tex] on the intervals created by the critical points and points of discontinuity.
1. Interval [tex]\((- \infty, -7)\)[/tex]:
Pick a test point, say [tex]\( x = -8 \)[/tex]:
[tex]\[ f'(-8) = \frac{-14(-8)}{((-8)^2 - 49)^2} > 0 \][/tex]
2. Interval [tex]\((-7, 0)\)[/tex]:
Pick a test point, say [tex]\( x = -1 \)[/tex]:
[tex]\[ f'(-1) = \frac{-14(-1)}{((-1)^2 - 49)^2} > 0 \][/tex]
3. Interval [tex]\((0, 7)\)[/tex]:
Pick a test point, say [tex]\( x = 1 \)[/tex]:
[tex]\[ f'(1) = \frac{-14(1)}{((1)^2 - 49)^2} < 0 \][/tex]
4. Interval [tex]\((7, \infty)\)[/tex]:
Pick a test point, say [tex]\( x = 8 \)[/tex]:
[tex]\[ f'(8) = \frac{-14(8)}{((8)^2 - 49)^2} < 0 \][/tex]
From the test points:
- [tex]\( f(x) \)[/tex] is increasing on the intervals [tex]\((- \infty, -7) \cup (-7, 0)\)[/tex].
Increasing: [tex]\( (- \infty, -7) U (-7, 0) \)[/tex]
### (C) Use interval notation to indicate where [tex]\( f(x) \)[/tex] is decreasing
From the analysis above:
- [tex]\( f(x) \)[/tex] is decreasing on the intervals [tex]\( (0, 7) \cup (7, \infty) \)[/tex].
Decreasing: [tex]\( (0, 7) U (7, \infty) \)[/tex]
### (D) List the [tex]\( x \)[/tex]-coordinates of all local maxima
A local maximum would occur where [tex]\( f'(x) \)[/tex] changes from positive to negative. From the intervals above and the nature of [tex]\( f'(x) \)[/tex]:
There is no point where [tex]\( f'(x) changes from positive to negative. Local maxima: \( NONE \)[/tex]
### (E) List the [tex]\( x \)[/tex]-coordinates of all local minima
A local minimum would occur where [tex]\( f'(x) \)[/tex] changes from negative to positive.
The only critical point is [tex]\( x = 0 \)[/tex], and we observed:
[tex]\( f'(x) \)[/tex] changes from positive to negative, indicating no local minima at [tex]\( x = 0 \)[/tex].
Local minima: [tex]\( NONE \)[/tex]
