Math / Calculus

QuestionName: $\qquad$ Part 1. Multiple Choice Graphing Calculator Permitted. Select the correct answer for each problem. Be sure to fill out the answer sheet on the back of the test. (a) $-\frac{9}{2}$ (b) -4 (c) 4 (d) $\frac{9}{2}$

Studdy Solution

STEP 1

What is this asking? We need to find the mystery value of $k$ that makes our funky function continuous at $x = 1$ . Watch out! Don't forget that continuity means the left-hand limit, right-hand limit, and the function's value at $x = 1$ must all be the same!

STEP 2

1. Evaluate the left-hand limit.
2. Evaluate the right-hand limit.
3. Set the limits equal and solve for k.

STEP 3

For $x$ approaching $\textbf{1}$ from the left ( $x < 1$ ), we use the top part of our function: $f(x) = -2x^2 - 7$ .
Let's plug in $x = 1$ !

STEP 4

$-2(\textbf{1})^2 - 7 = -2 \cdot 1 - 7 = -2 - 7 = \textbf{-9}$ So, our left-hand limit is $\textbf{-9}$ .

STEP 5

For $x$ approaching $\textbf{1}$ from the right ( $x > 1$ ), we use the bottom part: $f(x) = -x^2 - 2kx$ .
Let's substitute $x = 1$ !

STEP 6

$-(\textbf{1})^2 - 2k(\textbf{1}) = -1 - 2k$ So, our right-hand limit is $-1 - 2k$ .

STEP 7

For $f(x)$ to be continuous at $x = 1$ , the left-hand limit and the right-hand limit must be equal.
That means: $-9 = -1 - 2k$

STEP 8

We add 1 to both sides of the equation: $-9 + 1 = -1 - 2k + 1$ $-8 = -2k$

STEP 9

We divide both sides by $-2$ : $\frac{-8}{-2} = \frac{-2k}{-2}$ $\textbf{4} = k$

STEP 10

The value of $k$ that makes $f(x)$ continuous at $x = 1$ is $k = \textbf{4}$ .

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QuestionName: $\qquad$ Part 1. Multiple Choice Graphing Calculator Permitted. Select the correct answer for each problem. Be sure to fill out the answer sheet on the back of the test. (a) $-\frac{9}{2}$ (b) -4 (c) 4 (d) $\frac{9}{2}$

Studdy Solution

STEP 1

What is this asking? We need to find the mystery value of $k$ that makes our funky function continuous at $x = 1$ . Watch out! Don't forget that continuity means the left-hand limit, right-hand limit, and the function's value at $x = 1$ must all be the same!

STEP 2

1. Evaluate the left-hand limit.
2. Evaluate the right-hand limit.
3. Set the limits equal and solve for k.

STEP 3

For $x$ approaching $\textbf{1}$ from the left ( $x < 1$ ), we use the top part of our function: $f(x) = -2x^2 - 7$ .
Let's plug in $x = 1$ !

STEP 4

$-2(\textbf{1})^2 - 7 = -2 \cdot 1 - 7 = -2 - 7 = \textbf{-9}$ So, our left-hand limit is $\textbf{-9}$ .

STEP 5

For $x$ approaching $\textbf{1}$ from the right ( $x > 1$ ), we use the bottom part: $f(x) = -x^2 - 2kx$ .
Let's substitute $x = 1$ !

STEP 6

$-(\textbf{1})^2 - 2k(\textbf{1}) = -1 - 2k$ So, our right-hand limit is $-1 - 2k$ .

STEP 7

For $f(x)$ to be continuous at $x = 1$ , the left-hand limit and the right-hand limit must be equal.
That means: $-9 = -1 - 2k$

STEP 8

We add 1 to both sides of the equation: $-9 + 1 = -1 - 2k + 1$ $-8 = -2k$

STEP 9

We divide both sides by $-2$ : $\frac{-8}{-2} = \frac{-2k}{-2}$ $\textbf{4} = k$

STEP 10

The value of $k$ that makes $f(x)$ continuous at $x = 1$ is $k = \textbf{4}$ .

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QuestionName: \qquad Part 1. Multiple Choice Graphing Calculator Permitted. Select the correct answer for each problem. Be sure to fill out the answer sheet on the back of the test. (a) −92-\frac{9}{2}−29​ (b) -4 (c) 4 (d) 92\frac{9}{2}29​

Studdy Solution

STEP 1

What is this asking? We need to find the mystery value of kkk that makes our funky function continuous at x=1x = 1x=1. Watch out! Don't forget that continuity means the left-hand limit, right-hand limit, and the function's value at x=1x = 1x=1 must all be the same!

STEP 2

1. Evaluate the left-hand limit.2. Evaluate the right-hand limit.3. Set the limits equal and solve for *k*.

STEP 3

For xxx approaching 1\textbf{1}1 from the left (x<1x < 1x<1), we use the top part of our function: f(x)=−2x2−7f(x) = -2x^2 - 7f(x)=−2x2−7.Let's **plug in** x=1x = 1x=1!

STEP 4

−2(1)2−7=−2⋅1−7=−2−7=-9-2(\textbf{1})^2 - 7 = -2 \cdot 1 - 7 = -2 - 7 = \textbf{-9}−2(1)2−7=−2⋅1−7=−2−7=-9 So, our **left-hand limit** is -9\textbf{-9}-9.

STEP 5

For xxx approaching 1\textbf{1}1 from the right (x>1x > 1x>1), we use the bottom part: f(x)=−x2−2kxf(x) = -x^2 - 2kxf(x)=−x2−2kx.Let's **substitute** x=1x = 1x=1!

STEP 6

−(1)2−2k(1)=−1−2k-(\textbf{1})^2 - 2k(\textbf{1}) = -1 - 2k−(1)2−2k(1)=−1−2k So, our **right-hand limit** is −1−2k-1 - 2k−1−2k.

STEP 7

For f(x)f(x)f(x) to be continuous at x=1x = 1x=1, the left-hand limit and the right-hand limit must be equal.That means: −9=−1−2k-9 = -1 - 2k−9=−1−2k

STEP 8

We **add 1** to both sides of the equation: −9+1=−1−2k+1-9 + 1 = -1 - 2k + 1−9+1=−1−2k+1 −8=−2k-8 = -2k−8=−2k

STEP 9

We **divide both sides** by −2-2−2: −8−2=−2k−2\frac{-8}{-2} = \frac{-2k}{-2}−2−8​=−2−2k​ 4=k\textbf{4} = k4=k

STEP 10

The value of kkk that makes f(x)f(x)f(x) continuous at x=1x = 1x=1 is k=4k = \textbf{4}k=4.

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QuestionName: $\qquad$ Part 1. Multiple Choice Graphing Calculator Permitted. Select the correct answer for each problem. Be sure to fill out the answer sheet on the back of the test. (a) $-\frac{9}{2}$ (b) -4 (c) 4 (d) $\frac{9}{2}$

What is this asking? We need to find the mystery value of $k$ that makes our funky function continuous at $x = 1$ . Watch out! Don't forget that continuity means the left-hand limit, right-hand limit, and the function's value at $x = 1$ must all be the same!

1. Evaluate the left-hand limit.
2. Evaluate the right-hand limit.
3. Set the limits equal and solve for k.

For $x$ approaching $\textbf{1}$ from the left ( $x < 1$ ), we use the top part of our function: $f(x) = -2x^2 - 7$ .
Let's plug in $x = 1$ !

$-2(\textbf{1})^2 - 7 = -2 \cdot 1 - 7 = -2 - 7 = \textbf{-9}$ So, our left-hand limit is $\textbf{-9}$ .

For $x$ approaching $\textbf{1}$ from the right ( $x > 1$ ), we use the bottom part: $f(x) = -x^2 - 2kx$ .
Let's substitute $x = 1$ !

$-(\textbf{1})^2 - 2k(\textbf{1}) = -1 - 2k$ So, our right-hand limit is $-1 - 2k$ .

For $f(x)$ to be continuous at $x = 1$ , the left-hand limit and the right-hand limit must be equal.
That means: $-9 = -1 - 2k$

We add 1 to both sides of the equation: $-9 + 1 = -1 - 2k + 1$ $-8 = -2k$

We divide both sides by $-2$ : $\frac{-8}{-2} = \frac{-2k}{-2}$ $\textbf{4} = k$

The value of $k$ that makes $f(x)$ continuous at $x = 1$ is $k = \textbf{4}$ .