Solved on Mar 03, 2024

Find the relative extrema of $f(x) = -4 - x + x^2$ . Identify the intervals where the function is increasing and decreasing, and sketch the graph.

STEP 1

Assumptions
1. The given function is $f(x) = -4 - x + x^2$ .
2. We are looking for relative extrema, which are points where the function changes from increasing to decreasing or vice versa.
3. To find relative extrema, we will use the first derivative test.
4. We will also determine the intervals of increase and decrease of the function.

STEP 2

Find the first derivative of the function $f(x)$ with respect to $x$ .
$f'(x) = \frac{d}{dx}(-4 - x + x^2)$

STEP 3

Use the power rule to differentiate each term.
$f'(x) = 0 - 1 + 2x$

STEP 4

Simplify the derivative.
$f'(x) = 2x - 1$

STEP 5

Find the critical points by setting the first derivative equal to zero.
$2x - 1 = 0$

STEP 6

Solve for $x$ to find the critical points.
$2x = 1$
$x = \frac{1}{2}$

STEP 7

Determine the nature of the critical point using the first derivative test. This involves checking the sign of the derivative before and after the critical point $x = \frac{1}{2}$ .

STEP 8

Choose test points to the left and right of $x = \frac{1}{2}$ . For example, $x = 0$ for the left and $x = 1$ for the right.

STEP 9

Evaluate $f'(x)$ at $x = 0$ .
$f'(0) = 2(0) - 1 = -1$

STEP 10

Evaluate $f'(x)$ at $x = 1$ .
$f'(1) = 2(1) - 1 = 1$

STEP 11

Since $f'(x)$ changes from negative to positive as $x$ passes through $\frac{1}{2}$ , there is a relative minimum at $x = \frac{1}{2}$ .

STEP 12

To find the $y$ -value of the relative minimum, plug $x = \frac{1}{2}$ into the original function $f(x)$ .
$f\left(\frac{1}{2}\right) = -4 - \frac{1}{2} + \left(\frac{1}{2}\right)^2$

STEP 13

Simplify the expression to find the $y$ -value.
$f\left(\frac{1}{2}\right) = -4 - \frac{1}{2} + \frac{1}{4}$
$f\left(\frac{1}{2}\right) = -4 - \frac{2}{4} + \frac{1}{4}$
$f\left(\frac{1}{2}\right) = -4 - \frac{1}{4}$
$f\left(\frac{1}{2}\right) = -\frac{16}{4} - \frac{1}{4}$
$f\left(\frac{1}{2}\right) = -\frac{17}{4}$

STEP 14

The relative minimum point is $\left(\frac{1}{2}, -\frac{17}{4}\right)$ .

STEP 15

Determine the intervals of increase and decrease. From the first derivative test, we know that:
- The function is increasing when $f'(x) > 0$ , which is for $x > \frac{1}{2}$ . - The function is decreasing when $f'(x) < 0$ , which is for $x < \frac{1}{2}$ .

STEP 16

Write the intervals of increase and decrease in interval notation.
- The function is increasing on the interval $(\frac{1}{2}, \infty)$ . - The function is decreasing on the interval $(-\infty, \frac{1}{2})$ .

STEP 17

To sketch the graph, we note the following:
- The function has a relative minimum at $\left(\frac{1}{2}, -\frac{17}{4}\right)$ . - The function is decreasing before $x = \frac{1}{2}$ and increasing after $x = \frac{1}{2}$ . - The graph is a parabola opening upwards because the coefficient of $x^2$ is positive.

STEP 18

Sketch the graph using the information from the previous steps.

STEP 19

Combine the results to answer the problem:
- The relative minimum point(s) is/are $\left(\frac{1}{2}, -\frac{17}{4}\right)$ and there are no relative maximum points. - The function $f(x)$ is increasing over the interval $(\frac{1}{2}, \infty)$ and decreasing over the interval $(-\infty, \frac{1}{2})$ .

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Find the relative extrema of $f(x) = -4 - x + x^2$ . Identify the intervals where the function is increasing and decreasing, and sketch the graph.

STEP 1

STEP 2

Find the first derivative of the function $f(x)$ with respect to $x$ .
$f'(x) = \frac{d}{dx}(-4 - x + x^2)$

STEP 3

Use the power rule to differentiate each term.
$f'(x) = 0 - 1 + 2x$

STEP 4

Simplify the derivative.
$f'(x) = 2x - 1$

STEP 5

Find the critical points by setting the first derivative equal to zero.
$2x - 1 = 0$

STEP 6

Solve for $x$ to find the critical points.
$2x = 1$
$x = \frac{1}{2}$

STEP 7

Determine the nature of the critical point using the first derivative test. This involves checking the sign of the derivative before and after the critical point $x = \frac{1}{2}$ .

STEP 8

Choose test points to the left and right of $x = \frac{1}{2}$ . For example, $x = 0$ for the left and $x = 1$ for the right.

STEP 9

Evaluate $f'(x)$ at $x = 0$ .
$f'(0) = 2(0) - 1 = -1$

STEP 10

Evaluate $f'(x)$ at $x = 1$ .
$f'(1) = 2(1) - 1 = 1$

STEP 11

Since $f'(x)$ changes from negative to positive as $x$ passes through $\frac{1}{2}$ , there is a relative minimum at $x = \frac{1}{2}$ .

STEP 12

To find the $y$ -value of the relative minimum, plug $x = \frac{1}{2}$ into the original function $f(x)$ .
$f\left(\frac{1}{2}\right) = -4 - \frac{1}{2} + \left(\frac{1}{2}\right)^2$

STEP 13

STEP 14

The relative minimum point is $\left(\frac{1}{2}, -\frac{17}{4}\right)$ .

STEP 15

Determine the intervals of increase and decrease. From the first derivative test, we know that:
- The function is increasing when $f'(x) > 0$ , which is for $x > \frac{1}{2}$ . - The function is decreasing when $f'(x) < 0$ , which is for $x < \frac{1}{2}$ .

STEP 16

Write the intervals of increase and decrease in interval notation.
- The function is increasing on the interval $(\frac{1}{2}, \infty)$ . - The function is decreasing on the interval $(-\infty, \frac{1}{2})$ .

STEP 17

STEP 18

Sketch the graph using the information from the previous steps.

STEP 19

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Find the relative extrema of f(x)=−4−x+x2f(x) = -4 - x + x^2f(x)=−4−x+x2. Identify the intervals where the function is increasing and decreasing, and sketch the graph.

STEP 1

STEP 2

Find the first derivative of the function f(x) f(x) f(x) with respect to x x x.f′(x)=ddx(−4−x+x2) f'(x) = \frac{d}{dx}(-4 - x + x^2) f′(x)=dxd​(−4−x+x2)

STEP 3

Use the power rule to differentiate each term.f′(x)=0−1+2x f'(x) = 0 - 1 + 2x f′(x)=0−1+2x

STEP 4

Simplify the derivative.f′(x)=2x−1 f'(x) = 2x - 1 f′(x)=2x−1

STEP 5

Find the critical points by setting the first derivative equal to zero.2x−1=0 2x - 1 = 0 2x−1=0

STEP 6

Solve for x x x to find the critical points.2x=1 2x = 1 2x=1x=12 x = \frac{1}{2} x=21​

STEP 7

Determine the nature of the critical point using the first derivative test. This involves checking the sign of the derivative before and after the critical point x=12 x = \frac{1}{2} x=21​.

STEP 8

Choose test points to the left and right of x=12 x = \frac{1}{2} x=21​. For example, x=0 x = 0 x=0 for the left and x=1 x = 1 x=1 for the right.

STEP 9

Evaluate f′(x) f'(x) f′(x) at x=0 x = 0 x=0.f′(0)=2(0)−1=−1 f'(0) = 2(0) - 1 = -1 f′(0)=2(0)−1=−1

STEP 10

Evaluate f′(x) f'(x) f′(x) at x=1 x = 1 x=1.f′(1)=2(1)−1=1 f'(1) = 2(1) - 1 = 1 f′(1)=2(1)−1=1

STEP 11

Since f′(x) f'(x) f′(x) changes from negative to positive as x x x passes through 12 \frac{1}{2} 21​, there is a relative minimum at x=12 x = \frac{1}{2} x=21​.

STEP 12

To find the y y y-value of the relative minimum, plug x=12 x = \frac{1}{2} x=21​ into the original function f(x) f(x) f(x).f(12)=−4−12+(12)2 f\left(\frac{1}{2}\right) = -4 - \frac{1}{2} + \left(\frac{1}{2}\right)^2 f(21​)=−4−21​+(21​)2

STEP 13

STEP 14

The relative minimum point is (12,−174) \left(\frac{1}{2}, -\frac{17}{4}\right) (21​,−417​).

STEP 15

STEP 16

Write the intervals of increase and decrease in interval notation.- The function is increasing on the interval (12,∞) (\frac{1}{2}, \infty) (21​,∞). - The function is decreasing on the interval (−∞,12) (-\infty, \frac{1}{2}) (−∞,21​).

STEP 17

STEP 18

Sketch the graph using the information from the previous steps.

STEP 19

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Find the relative extrema of $f(x) = -4 - x + x^2$ . Identify the intervals where the function is increasing and decreasing, and sketch the graph.

Find the first derivative of the function $f(x)$ with respect to $x$ .
$f'(x) = \frac{d}{dx}(-4 - x + x^2)$

Use the power rule to differentiate each term.
$f'(x) = 0 - 1 + 2x$

Simplify the derivative.
$f'(x) = 2x - 1$

Find the critical points by setting the first derivative equal to zero.
$2x - 1 = 0$

Solve for $x$ to find the critical points.
$2x = 1$
$x = \frac{1}{2}$

Determine the nature of the critical point using the first derivative test. This involves checking the sign of the derivative before and after the critical point $x = \frac{1}{2}$ .

Choose test points to the left and right of $x = \frac{1}{2}$ . For example, $x = 0$ for the left and $x = 1$ for the right.

Evaluate $f'(x)$ at $x = 0$ .
$f'(0) = 2(0) - 1 = -1$

Evaluate $f'(x)$ at $x = 1$ .
$f'(1) = 2(1) - 1 = 1$

Since $f'(x)$ changes from negative to positive as $x$ passes through $\frac{1}{2}$ , there is a relative minimum at $x = \frac{1}{2}$ .

To find the $y$ -value of the relative minimum, plug $x = \frac{1}{2}$ into the original function $f(x)$ .
$f\left(\frac{1}{2}\right) = -4 - \frac{1}{2} + \left(\frac{1}{2}\right)^2$

The relative minimum point is $\left(\frac{1}{2}, -\frac{17}{4}\right)$ .

Write the intervals of increase and decrease in interval notation.
- The function is increasing on the interval $(\frac{1}{2}, \infty)$ . - The function is decreasing on the interval $(-\infty, \frac{1}{2})$ .