Math / Calculus

QuestionLet $A(x)=\int_{0}^{x} f(t) d t$ , with $f(x)$ as in figure. $A(x)$ has a local minimum on $(0,6)$ at $x=$ $\square$ $A(x)$ has a local maximum on $(0,6)$ at $x=$ $\square$

Studdy Solution

STEP 1

1. $A(x) = \int_{0}^{x} f(t) \, dt$ is the integral of the function $f(x)$ from 0 to $x$ .
2. The function $f(x)$ is piecewise linear with key points given in the problem.
3. We are looking for local minima and maxima of $A(x)$ on the interval $(0, 6)$ .

STEP 2

1. Understand the relationship between $A(x)$ and $f(x)$ .
2. Analyze the behavior of $f(x)$ to determine critical points of $A(x)$ .
3. Determine where $A(x)$ has local minima and maxima.

STEP 3

The derivative of $A(x)$ with respect to $x$ is $A'(x) = f(x)$ . This means that the critical points of $A(x)$ occur where $f(x) = 0$ .

STEP 4

Identify where $f(x) = 0$ based on the graph: - $f(x)$ crosses the x-axis at $x = 2$ and $x = 6$ .

STEP 5

Since $f(x)$ changes sign at these points, they are potential locations for local minima or maxima of $A(x)$ .

STEP 6

Evaluate the behavior of $f(x)$ around these points: - From $x = 0$ to $x = 2$ , $f(x)$ is negative, so $A(x)$ is decreasing. - From $x = 2$ to $x = 4$ , $f(x)$ is positive, so $A(x)$ is increasing. - From $x = 4$ to $x = 6$ , $f(x)$ is negative, so $A(x)$ is decreasing.

STEP 7

Determine local minima and maxima: - At $x = 2$ , $A(x)$ changes from decreasing to increasing, indicating a local minimum. - At $x = 4$ , $A(x)$ changes from increasing to decreasing, indicating a local maximum.
The answers are: $A(x) \text{ has a local minimum at } x = 2$ $A(x) \text{ has a local maximum at } x = 4$

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QuestionLet $A(x)=\int_{0}^{x} f(t) d t$ , with $f(x)$ as in figure. $A(x)$ has a local minimum on $(0,6)$ at $x=$ $\square$ $A(x)$ has a local maximum on $(0,6)$ at $x=$ $\square$

Studdy Solution

STEP 1

1. $A(x) = \int_{0}^{x} f(t) \, dt$ is the integral of the function $f(x)$ from 0 to $x$ .
2. The function $f(x)$ is piecewise linear with key points given in the problem.
3. We are looking for local minima and maxima of $A(x)$ on the interval $(0, 6)$ .

STEP 2

1. Understand the relationship between $A(x)$ and $f(x)$ .
2. Analyze the behavior of $f(x)$ to determine critical points of $A(x)$ .
3. Determine where $A(x)$ has local minima and maxima.

STEP 3

The derivative of $A(x)$ with respect to $x$ is $A'(x) = f(x)$ . This means that the critical points of $A(x)$ occur where $f(x) = 0$ .

STEP 4

Identify where $f(x) = 0$ based on the graph: - $f(x)$ crosses the x-axis at $x = 2$ and $x = 6$ .

STEP 5

Since $f(x)$ changes sign at these points, they are potential locations for local minima or maxima of $A(x)$ .

STEP 6

Evaluate the behavior of $f(x)$ around these points: - From $x = 0$ to $x = 2$ , $f(x)$ is negative, so $A(x)$ is decreasing. - From $x = 2$ to $x = 4$ , $f(x)$ is positive, so $A(x)$ is increasing. - From $x = 4$ to $x = 6$ , $f(x)$ is negative, so $A(x)$ is decreasing.

STEP 7

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QuestionLet A(x)=∫0xf(t)dtA(x)=\int_{0}^{x} f(t) d tA(x)=∫0x​f(t)dt, with f(x)f(x)f(x) as in figure. A(x)A(x)A(x) has a local minimum on (0,6)(0,6)(0,6) at x=x=x= □\square□ A(x)A(x)A(x) has a local maximum on (0,6)(0,6)(0,6) at x=x=x= □\square□

Studdy Solution

STEP 1

STEP 2

1. Understand the relationship between A(x) A(x) A(x) and f(x) f(x) f(x).2. Analyze the behavior of f(x) f(x) f(x) to determine critical points of A(x) A(x) A(x).3. Determine where A(x) A(x) A(x) has local minima and maxima.

STEP 3

The derivative of A(x) A(x) A(x) with respect to x x x is A′(x)=f(x) A'(x) = f(x) A′(x)=f(x). This means that the critical points of A(x) A(x) A(x) occur where f(x)=0 f(x) = 0 f(x)=0.

STEP 4

Identify where f(x)=0 f(x) = 0 f(x)=0 based on the graph: - f(x) f(x) f(x) crosses the x-axis at x=2 x = 2 x=2 and x=6 x = 6 x=6.

STEP 5

Since f(x) f(x) f(x) changes sign at these points, they are potential locations for local minima or maxima of A(x) A(x) A(x).

STEP 6

STEP 7

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QuestionLet $A(x)=\int_{0}^{x} f(t) d t$ , with $f(x)$ as in figure. $A(x)$ has a local minimum on $(0,6)$ at $x=$ $\square$ $A(x)$ has a local maximum on $(0,6)$ at $x=$ $\square$

1. Understand the relationship between $A(x)$ and $f(x)$ .
2. Analyze the behavior of $f(x)$ to determine critical points of $A(x)$ .
3. Determine where $A(x)$ has local minima and maxima.

The derivative of $A(x)$ with respect to $x$ is $A'(x) = f(x)$ . This means that the critical points of $A(x)$ occur where $f(x) = 0$ .

Identify where $f(x) = 0$ based on the graph: - $f(x)$ crosses the x-axis at $x = 2$ and $x = 6$ .

Since $f(x)$ changes sign at these points, they are potential locations for local minima or maxima of $A(x)$ .