Question Find intervals where $f(x) = \sqrt{-2x^2 - 2x}$ is continuous. If multiple intervals, separate with $\cup$ or comma. If not continuous, output $\varnothing$ .

Studdy Solution

STEP 1

Assumptions
1. The function given is $f(x)=\sqrt{-2 x^{2}-2 x}$ .
2. A function is continuous at a point if it is defined at that point and the limit of the function as it approaches that point exists and is equal to the function's value at that point.
3. A square root function, $\sqrt{x}$ , is defined for $x \geq 0$ .
4. We need to find the interval(s) where the argument of the square root, $-2 x^{2}-2 x$ , is greater than or equal to zero.

STEP 2

First, we need to find the domain of the function inside the square root, which is $-2 x^{2}-2 x$ . We want to find the values of $x$ for which $-2 x^{2}-2 x \geq 0$ .

STEP 3

Factor out the common factor of $-2$ from the quadratic expression.
$-2 x^{2}-2 x = -2(x^{2}+x)$

STEP 4

Factor the quadratic expression $x^{2}+x$ .
$x^{2}+x = x(x+1)$

STEP 5

Now we have the inequality $-2(x)(x+1) \geq 0$ . To solve this inequality, we need to find the zeros of the expression and determine the sign of the expression in the intervals defined by these zeros.

STEP 6

Set the factored expression equal to zero to find the zeros.
$-2(x)(x+1) = 0$

STEP 7

Solve for $x$ .
$x = 0 \quad \text{or} \quad x = -1$

STEP 8

Use these zeros to determine the intervals to test. The intervals are $(-\infty, -1)$ , $(-1, 0)$ , and $(0, +\infty)$ .

STEP 9

Choose a test point from each interval and plug it into the factored inequality to determine the sign of the expression in that interval.
For $(-\infty, -1)$ , choose $x = -2$ : $-2(-2)(-2+1) = -2(4)(-1) = 8 > 0$ However, since we have a negative coefficient $-2$ outside the parentheses, the expression will be negative in this interval.
For $(-1, 0)$ , choose $x = -\frac{1}{2}$ : $-2\left(-\frac{1}{2}\right)\left(-\frac{1}{2}+1\right) = -2\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right) = \frac{1}{2} > 0$ In this interval, the expression is positive.
For $(0, +\infty)$ , choose $x = 1$ : $-2(1)(1+1) = -2(2) = -4 < 0$ In this interval, the expression is negative.

STEP 10

From the test points, we can see that the expression $-2 x^{2}-2 x$ is non-negative in the interval $(-1, 0)$ .

STEP 11

Therefore, the function $f(x)=\sqrt{-2 x^{2}-2 x}$ is continuous over the interval $(-1, 0)$ .
The interval(s) over which the function is continuous is $(-1, 0)$ .

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Question Find intervals where $f(x) = \sqrt{-2x^2 - 2x}$ is continuous. If multiple intervals, separate with $\cup$ or comma. If not continuous, output $\varnothing$ .

Studdy Solution

STEP 1

STEP 2

First, we need to find the domain of the function inside the square root, which is $-2 x^{2}-2 x$ . We want to find the values of $x$ for which $-2 x^{2}-2 x \geq 0$ .

STEP 3

Factor out the common factor of $-2$ from the quadratic expression.
$-2 x^{2}-2 x = -2(x^{2}+x)$

STEP 4

Factor the quadratic expression $x^{2}+x$ .
$x^{2}+x = x(x+1)$

STEP 5

Now we have the inequality $-2(x)(x+1) \geq 0$ . To solve this inequality, we need to find the zeros of the expression and determine the sign of the expression in the intervals defined by these zeros.

STEP 6

Set the factored expression equal to zero to find the zeros.
$-2(x)(x+1) = 0$

STEP 7

Solve for $x$ .
$x = 0 \quad \text{or} \quad x = -1$

STEP 8

Use these zeros to determine the intervals to test. The intervals are $(-\infty, -1)$ , $(-1, 0)$ , and $(0, +\infty)$ .

STEP 9

STEP 10

From the test points, we can see that the expression $-2 x^{2}-2 x$ is non-negative in the interval $(-1, 0)$ .

STEP 11

Therefore, the function $f(x)=\sqrt{-2 x^{2}-2 x}$ is continuous over the interval $(-1, 0)$ .
The interval(s) over which the function is continuous is $(-1, 0)$ .

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Question Find intervals where f(x)=−2x2−2xf(x) = \sqrt{-2x^2 - 2x}f(x)=−2x2−2x​ is continuous. If multiple intervals, separate with ∪\cup∪ or comma. If not continuous, output ∅\varnothing∅.

Studdy Solution

STEP 1

STEP 2

First, we need to find the domain of the function inside the square root, which is −2x2−2x-2 x^{2}-2 x−2x2−2x. We want to find the values of xxx for which −2x2−2x≥0-2 x^{2}-2 x \geq 0−2x2−2x≥0.

STEP 3

Factor out the common factor of −2-2−2 from the quadratic expression.−2x2−2x=−2(x2+x)-2 x^{2}-2 x = -2(x^{2}+x)−2x2−2x=−2(x2+x)

STEP 4

Factor the quadratic expression x2+xx^{2}+xx2+x.x2+x=x(x+1)x^{2}+x = x(x+1)x2+x=x(x+1)

STEP 5

Now we have the inequality −2(x)(x+1)≥0-2(x)(x+1) \geq 0−2(x)(x+1)≥0. To solve this inequality, we need to find the zeros of the expression and determine the sign of the expression in the intervals defined by these zeros.

STEP 6

Set the factored expression equal to zero to find the zeros.−2(x)(x+1)=0-2(x)(x+1) = 0−2(x)(x+1)=0

STEP 7

Solve for xxx.x=0orx=−1x = 0 \quad \text{or} \quad x = -1x=0orx=−1

STEP 8

Use these zeros to determine the intervals to test. The intervals are (−∞,−1)(-\infty, -1)(−∞,−1), (−1,0)(-1, 0)(−1,0), and (0,+∞)(0, +\infty)(0,+∞).

STEP 9

STEP 10

From the test points, we can see that the expression −2x2−2x-2 x^{2}-2 x−2x2−2x is non-negative in the interval (−1,0)(-1, 0)(−1,0).

STEP 11

Therefore, the function f(x)=−2x2−2xf(x)=\sqrt{-2 x^{2}-2 x}f(x)=−2x2−2x​ is continuous over the interval (−1,0)(-1, 0)(−1,0).The interval(s) over which the function is continuous is (−1,0)(-1, 0)(−1,0).

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Question Find intervals where $f(x) = \sqrt{-2x^2 - 2x}$ is continuous. If multiple intervals, separate with $\cup$ or comma. If not continuous, output $\varnothing$ .

First, we need to find the domain of the function inside the square root, which is $-2 x^{2}-2 x$ . We want to find the values of $x$ for which $-2 x^{2}-2 x \geq 0$ .

Factor out the common factor of $-2$ from the quadratic expression.
$-2 x^{2}-2 x = -2(x^{2}+x)$

Factor the quadratic expression $x^{2}+x$ .
$x^{2}+x = x(x+1)$

Now we have the inequality $-2(x)(x+1) \geq 0$ . To solve this inequality, we need to find the zeros of the expression and determine the sign of the expression in the intervals defined by these zeros.

Set the factored expression equal to zero to find the zeros.
$-2(x)(x+1) = 0$

Solve for $x$ .
$x = 0 \quad \text{or} \quad x = -1$

Use these zeros to determine the intervals to test. The intervals are $(-\infty, -1)$ , $(-1, 0)$ , and $(0, +\infty)$ .

From the test points, we can see that the expression $-2 x^{2}-2 x$ is non-negative in the interval $(-1, 0)$ .

Therefore, the function $f(x)=\sqrt{-2 x^{2}-2 x}$ is continuous over the interval $(-1, 0)$ .
The interval(s) over which the function is continuous is $(-1, 0)$ .