QuestionFind the value of $c$ for the piecewise function $f(x)=\{2x+c, x \leq 1; x^2+3, x>1\}$ to be continuous.

Studdy Solution

STEP 1

Assumptions1. The function is a piecewise function defined as $f(x)=\left\{\begin{array}{ll} x+c & x \leq1 \\ x^{}+3 & x>1\end{array}\right.$ . We need to find the value of $c$ which makes the function continuous.

STEP 2

For a function to be continuous at a point, the left-hand limit, right-hand limit, and the function value at that point must be equal. As the function changes definition at $x=1$ , we need to check the continuity at $x=1$ .

STEP 3

First, let's calculate the left-hand limit of the function at $x=1$ . This is done by substituting $x=1$ into the first part of the piecewise function.
$\lim{{x \to1^-}} f(x) =2(1) + c =2 + c$

STEP 4

Next, let's calculate the right-hand limit of the function at $x=1$ . This is done by substituting $x=1$ into the second part of the piecewise function.
$\lim{{x \to1^+}} f(x) = (1)^2 +3 =1 +3 =4$

STEP 5

For the function to be continuous at $x=1$ , the left-hand limit must equal the right-hand limit. Therefore, we set the two limits equal to each other and solve for $c$ .
$2 + c =4$

STEP 6

Subtract2 from both sides to solve for $c$ .
$c =4 -2 =2$ The value of $c$ which makes the function continuous is $2$ .

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QuestionFind the value of $c$ for the piecewise function $f(x)=\{2x+c, x \leq 1; x^2+3, x>1\}$ to be continuous.

Studdy Solution

STEP 1

Assumptions1. The function is a piecewise function defined as $f(x)=\left\{\begin{array}{ll} x+c & x \leq1 \\ x^{}+3 & x>1\end{array}\right.$ . We need to find the value of $c$ which makes the function continuous.

STEP 2

For a function to be continuous at a point, the left-hand limit, right-hand limit, and the function value at that point must be equal. As the function changes definition at $x=1$ , we need to check the continuity at $x=1$ .

STEP 3

First, let's calculate the left-hand limit of the function at $x=1$ . This is done by substituting $x=1$ into the first part of the piecewise function.
$\lim{{x \to1^-}} f(x) =2(1) + c =2 + c$

STEP 4

Next, let's calculate the right-hand limit of the function at $x=1$ . This is done by substituting $x=1$ into the second part of the piecewise function.
$\lim{{x \to1^+}} f(x) = (1)^2 +3 =1 +3 =4$

STEP 5

For the function to be continuous at $x=1$ , the left-hand limit must equal the right-hand limit. Therefore, we set the two limits equal to each other and solve for $c$ .
$2 + c =4$

STEP 6

Subtract2 from both sides to solve for $c$ .
$c =4 -2 =2$ The value of $c$ which makes the function continuous is $2$ .

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QuestionFind the value of ccc for the piecewise function f(x)={2x+c,x≤1;x2+3,x>1}f(x)=\{2x+c, x \leq 1; x^2+3, x>1\}f(x)={2x+c,x≤1;x2+3,x>1} to be continuous.

Studdy Solution

STEP 1

Assumptions1. The function is a piecewise function defined asf(x)={x+cx≤1x+3x>1f(x)=\left\{\begin{array}{ll} x+c & x \leq1 \\ x^{}+3 & x>1\end{array}\right. f(x)={x+cx+3​x≤1x>1​. We need to find the value of ccc which makes the function continuous.

STEP 2

For a function to be continuous at a point, the left-hand limit, right-hand limit, and the function value at that point must be equal. As the function changes definition at x=1x=1x=1, we need to check the continuity at x=1x=1x=1.

STEP 3

First, let's calculate the left-hand limit of the function at x=1x=1x=1. This is done by substituting x=1x=1x=1 into the first part of the piecewise function.lim⁡x→1−f(x)=2(1)+c=2+c\lim{{x \to1^-}} f(x) =2(1) + c =2 + climx→1−f(x)=2(1)+c=2+c

STEP 4

Next, let's calculate the right-hand limit of the function at x=1x=1x=1. This is done by substituting x=1x=1x=1 into the second part of the piecewise function.lim⁡x→1+f(x)=(1)2+3=1+3=4\lim{{x \to1^+}} f(x) = (1)^2 +3 =1 +3 =4limx→1+f(x)=(1)2+3=1+3=4

STEP 5

For the function to be continuous at x=1x=1x=1, the left-hand limit must equal the right-hand limit. Therefore, we set the two limits equal to each other and solve for ccc.2+c=42 + c =42+c=4

STEP 6

Subtract2 from both sides to solve for ccc.c=4−2=2c =4 -2 =2c=4−2=2The value of ccc which makes the function continuous is 222.

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QuestionFind the value of $c$ for the piecewise function $f(x)=\{2x+c, x \leq 1; x^2+3, x>1\}$ to be continuous.

Assumptions1. The function is a piecewise function defined as $f(x)=\left\{\begin{array}{ll} x+c & x \leq1 \\ x^{}+3 & x>1\end{array}\right.$ . We need to find the value of $c$ which makes the function continuous.

For a function to be continuous at a point, the left-hand limit, right-hand limit, and the function value at that point must be equal. As the function changes definition at $x=1$ , we need to check the continuity at $x=1$ .

First, let's calculate the left-hand limit of the function at $x=1$ . This is done by substituting $x=1$ into the first part of the piecewise function.
$\lim{{x \to1^-}} f(x) =2(1) + c =2 + c$

Next, let's calculate the right-hand limit of the function at $x=1$ . This is done by substituting $x=1$ into the second part of the piecewise function.
$\lim{{x \to1^+}} f(x) = (1)^2 +3 =1 +3 =4$

For the function to be continuous at $x=1$ , the left-hand limit must equal the right-hand limit. Therefore, we set the two limits equal to each other and solve for $c$ .
$2 + c =4$

Subtract2 from both sides to solve for $c$ .
$c =4 -2 =2$ The value of $c$ which makes the function continuous is $2$ .