QuestionFind the value of $a$ for which $f(x)$ is continuous at all $x$ :
$f(x)=\begin{cases} x^{2}-99, & x<11 \\ 2 a x, & x \geq 11 \end{cases}$
Options: A. $a=$ B. No solution.

Studdy Solution

STEP 1

Assumptions1. The function $f(x)$ is given by $f(x) = x^ -99$ for $x <11$ and $f(x) =ax$ for $x \geq11$ . . We are asked to find the value of $a$ that makes the function continuous at every $x$ .

STEP 2

A function is continuous at a point if the limit from the left equals the limit from the right at that point. In this case, we need the function to be continuous at $x =11$ . So, we need to set the limit from the left equal to the limit from the right at $x =11$ .
$\lim{{x \to11^-}} (x^2 -99) = \lim{{x \to11^+}} (2ax)$

STEP 3

We can calculate the limit from the left at $x =11$ by substituting $x =11$ into the function $x^2 -99$ .
$\lim{{x \to11^-}} (x^2 -99) =11^2 -99$

STEP 4

Calculate the left limit.
$\lim{{x \to11^-}} (x^2 -99) =11^2 -99 =22$

STEP 5

We can calculate the limit from the right at $x =11$ by substituting $x =11$ into the function $2ax$ .
$\lim{{x \to11^+}} (2ax) =2a \cdot11$

STEP 6

Now, we can set the two limits equal to each other and solve for $a$ .
$22 =2a \cdot11$

STEP 7

olve for $a$ .
$a = \frac{22}{2 \cdot11}$

STEP 8

Calculate the value of $a$ .
$a = \frac{22}{22} =1$ The value of $a$ that makes the function continuous at every $x$ is $1$ .

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QuestionFind the value of $a$ for which $f(x)$ is continuous at all $x$ :
$f(x)=\begin{cases} x^{2}-99, & x<11 \\ 2 a x, & x \geq 11 \end{cases}$
Options: A. $a=$ B. No solution.

Studdy Solution

STEP 1

Assumptions1. The function $f(x)$ is given by $f(x) = x^ -99$ for $x <11$ and $f(x) =ax$ for $x \geq11$ . . We are asked to find the value of $a$ that makes the function continuous at every $x$ .

STEP 2

STEP 3

We can calculate the limit from the left at $x =11$ by substituting $x =11$ into the function $x^2 -99$ .
$\lim{{x \to11^-}} (x^2 -99) =11^2 -99$

STEP 4

Calculate the left limit.
$\lim{{x \to11^-}} (x^2 -99) =11^2 -99 =22$

STEP 5

We can calculate the limit from the right at $x =11$ by substituting $x =11$ into the function $2ax$ .
$\lim{{x \to11^+}} (2ax) =2a \cdot11$

STEP 6

Now, we can set the two limits equal to each other and solve for $a$ .
$22 =2a \cdot11$

STEP 7

olve for $a$ .
$a = \frac{22}{2 \cdot11}$

STEP 8

Calculate the value of $a$ .
$a = \frac{22}{22} =1$ The value of $a$ that makes the function continuous at every $x$ is $1$ .

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QuestionFind the value of aaa for which f(x)f(x)f(x) is continuous at all xxx:f(x)={x2−99,x<112ax,x≥11 f(x)=\begin{cases} x^{2}-99, & x<11 \\ 2 a x, & x \geq 11 \end{cases} f(x)={x2−99,2ax,​x<11x≥11​Options: A. a=a=a= B. No solution.

Studdy Solution

STEP 1

Assumptions1. The function f(x)f(x)f(x) is given by f(x)=x−99f(x) = x^ -99f(x)=x−99 for x<11x <11x<11 and f(x)=axf(x) =axf(x)=ax for x≥11x \geq11x≥11. . We are asked to find the value of aaa that makes the function continuous at every xxx.

STEP 2

STEP 3

We can calculate the limit from the left at x=11x =11x=11 by substituting x=11x =11x=11 into the function x2−99x^2 -99x2−99.lim⁡x→11−(x2−99)=112−99\lim{{x \to11^-}} (x^2 -99) =11^2 -99limx→11−(x2−99)=112−99

STEP 4

Calculate the left limit.lim⁡x→11−(x2−99)=112−99=22\lim{{x \to11^-}} (x^2 -99) =11^2 -99 =22limx→11−(x2−99)=112−99=22

STEP 5

We can calculate the limit from the right at x=11x =11x=11 by substituting x=11x =11x=11 into the function 2ax2ax2ax.lim⁡x→11+(2ax)=2a⋅11\lim{{x \to11^+}} (2ax) =2a \cdot11limx→11+(2ax)=2a⋅11

STEP 6

Now, we can set the two limits equal to each other and solve for aaa.22=2a⋅1122 =2a \cdot1122=2a⋅11

STEP 7

olve for aaa.a=222⋅11a = \frac{22}{2 \cdot11}a=2⋅1122​

STEP 8

Calculate the value of aaa.a=2222=1a = \frac{22}{22} =1a=2222​=1The value of aaa that makes the function continuous at every xxx is 111.

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QuestionFind the value of $a$ for which $f(x)$ is continuous at all $x$ :
$f(x)=\begin{cases} x^{2}-99, & x<11 \\ 2 a x, & x \geq 11 \end{cases}$
Options: A. $a=$ B. No solution.

Assumptions1. The function $f(x)$ is given by $f(x) = x^ -99$ for $x <11$ and $f(x) =ax$ for $x \geq11$ . . We are asked to find the value of $a$ that makes the function continuous at every $x$ .

We can calculate the limit from the left at $x =11$ by substituting $x =11$ into the function $x^2 -99$ .
$\lim{{x \to11^-}} (x^2 -99) =11^2 -99$

Calculate the left limit.
$\lim{{x \to11^-}} (x^2 -99) =11^2 -99 =22$

We can calculate the limit from the right at $x =11$ by substituting $x =11$ into the function $2ax$ .
$\lim{{x \to11^+}} (2ax) =2a \cdot11$

Now, we can set the two limits equal to each other and solve for $a$ .
$22 =2a \cdot11$

olve for $a$ .
$a = \frac{22}{2 \cdot11}$

Calculate the value of $a$ .
$a = \frac{22}{22} =1$ The value of $a$ that makes the function continuous at every $x$ is $1$ .