Math / Calculus

Question32 Mark for Review
Let $f$ be the function with derivative given by $f^{\prime}(x)=\sin x+x \cos x$ for $0<x \leq \pi$ . On which of the following intervals is $f$ increasing? (A) $[0,1.077]$ only (B) $[0,2.029]$ (C) (D) $[2.029, \pi]$ only

Studdy Solution

STEP 1

1. The function $f$ is differentiable on the interval $0 < x \leq \pi$ .
2. $f$ is increasing where its derivative $f'(x)$ is positive.
3. We need to analyze the sign of $f'(x) = \sin x + x \cos x$ over the given intervals.

STEP 2

1. Analyze the expression $f'(x) = \sin x + x \cos x$ .
2. Determine where $f'(x) > 0$ .
3. Evaluate the sign of $f'(x)$ over the given intervals.
4. Identify the interval(s) where $f$ is increasing.

STEP 3

The expression for the derivative is given as:
$f'(x) = \sin x + x \cos x$

STEP 4

To determine where $f'(x) > 0$ , we need to solve:
$\sin x + x \cos x > 0$
This inequality will help us find where the function $f$ is increasing.

STEP 5

We will analyze $f'(x)$ at critical points and endpoints of the intervals. First, check the endpoints $x = 0$ and $x = \pi$ :
- At $x = 0$ : $f'(0) = \sin(0) + 0 \cdot \cos(0) = 0$
- At $x = \pi$ : $f'(\pi) = \sin(\pi) + \pi \cdot \cos(\pi) = 0 - \pi = -\pi$

STEP 6

Next, we need to evaluate $f'(x)$ at the critical points within the intervals, particularly around $x = 1.077$ and $x = 2.029$ :
- At $x = 1.077$ : Calculate $f'(1.077) = \sin(1.077) + 1.077 \cdot \cos(1.077)$ and check if it is positive.
- At $x = 2.029$ : Calculate $f'(2.029) = \sin(2.029) + 2.029 \cdot \cos(2.029)$ and check if it is positive.

STEP 7

Evaluate the sign of $f'(x)$ over the intervals:
- For $[0, 1.077]$ , check if $f'(x) > 0$ for all $x$ in this interval. - For $[0, 2.029]$ , check if $f'(x) > 0$ for all $x$ in this interval. - For $[2.029, \pi]$ , check if $f'(x) > 0$ for all $x$ in this interval.

STEP 8

Based on the evaluations, identify the interval(s) where $f'(x) > 0$ and thus $f$ is increasing.
- If $f'(x) > 0$ on $[0, 1.077]$ , then $f$ is increasing on this interval. - If $f'(x) > 0$ on $[0, 2.029]$ , then $f$ is increasing on this interval. - If $f'(x) > 0$ on $[2.029, \pi]$ , then $f$ is increasing on this interval.
The correct interval where $f$ is increasing is $\boxed{[0, 2.029]}$ .

Was this helpful?

Question32 Mark for Review
Let $f$ be the function with derivative given by $f^{\prime}(x)=\sin x+x \cos x$ for $0<x \leq \pi$ . On which of the following intervals is $f$ increasing? (A) $[0,1.077]$ only (B) $[0,2.029]$ (C) (D) $[2.029, \pi]$ only

Studdy Solution

STEP 1

1. The function $f$ is differentiable on the interval $0 < x \leq \pi$ .
2. $f$ is increasing where its derivative $f'(x)$ is positive.
3. We need to analyze the sign of $f'(x) = \sin x + x \cos x$ over the given intervals.

STEP 2

1. Analyze the expression $f'(x) = \sin x + x \cos x$ .
2. Determine where $f'(x) > 0$ .
3. Evaluate the sign of $f'(x)$ over the given intervals.
4. Identify the interval(s) where $f$ is increasing.

STEP 3

The expression for the derivative is given as:
$f'(x) = \sin x + x \cos x$

STEP 4

To determine where $f'(x) > 0$ , we need to solve:
$\sin x + x \cos x > 0$
This inequality will help us find where the function $f$ is increasing.

STEP 5

We will analyze $f'(x)$ at critical points and endpoints of the intervals. First, check the endpoints $x = 0$ and $x = \pi$ :
- At $x = 0$ : $f'(0) = \sin(0) + 0 \cdot \cos(0) = 0$
- At $x = \pi$ : $f'(\pi) = \sin(\pi) + \pi \cdot \cos(\pi) = 0 - \pi = -\pi$

STEP 6

STEP 7

Evaluate the sign of $f'(x)$ over the intervals:
- For $[0, 1.077]$ , check if $f'(x) > 0$ for all $x$ in this interval. - For $[0, 2.029]$ , check if $f'(x) > 0$ for all $x$ in this interval. - For $[2.029, \pi]$ , check if $f'(x) > 0$ for all $x$ in this interval.

STEP 8

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

1. The function f f f is differentiable on the interval 0<x≤π 0 < x \leq \pi 0<x≤π.2. f f f is increasing where its derivative f′(x) f'(x) f′(x) is positive.3. We need to analyze the sign of f′(x)=sin⁡x+xcos⁡x f'(x) = \sin x + x \cos x f′(x)=sinx+xcosx over the given intervals.

STEP 2

1. Analyze the expression f′(x)=sin⁡x+xcos⁡x f'(x) = \sin x + x \cos x f′(x)=sinx+xcosx.2. Determine where f′(x)>0 f'(x) > 0 f′(x)>0.3. Evaluate the sign of f′(x) f'(x) f′(x) over the given intervals.4. Identify the interval(s) where f f f is increasing.

STEP 3

The expression for the derivative is given as:f′(x)=sin⁡x+xcos⁡x f'(x) = \sin x + x \cos x f′(x)=sinx+xcosx

STEP 4

To determine where f′(x)>0 f'(x) > 0 f′(x)>0, we need to solve:sin⁡x+xcos⁡x>0 \sin x + x \cos x > 0 sinx+xcosx>0This inequality will help us find where the function f f f is increasing.

STEP 5

STEP 6

STEP 7

STEP 8

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. The function $f$ is differentiable on the interval $0 < x \leq \pi$ .
2. $f$ is increasing where its derivative $f'(x)$ is positive.
3. We need to analyze the sign of $f'(x) = \sin x + x \cos x$ over the given intervals.

1. Analyze the expression $f'(x) = \sin x + x \cos x$ .
2. Determine where $f'(x) > 0$ .
3. Evaluate the sign of $f'(x)$ over the given intervals.
4. Identify the interval(s) where $f$ is increasing.

The expression for the derivative is given as:
$f'(x) = \sin x + x \cos x$

To determine where $f'(x) > 0$ , we need to solve:
$\sin x + x \cos x > 0$
This inequality will help us find where the function $f$ is increasing.