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question 9
0/1 pt 100 19 details
let $f(x)=\

$$\begin{cases}mx - 12 & \\text{if } x < -8 \\\\ x^2 + 2x - 4 & \\text{if } x \\geq -8\\end{cases}$$

$
if $f(x)$ is a function which is continuous everywhere, then we must have
$m = \square$
question help: video
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question 10
0/1 pt 100 19 details
find $c$ such that the function
$f(x)=\

$$\begin{cases}x^2 - 6 & x \\leq c \\\\ 4x - 10 & x > c\\end{cases}$$

$
is continuous everywhere.
$c = \square$
question help: video

Explanation:

Question 9

Step1: Set functions equal at $x=-8$

For continuity, $mx-12 = x^2+2x-4$ at $x=-8$.

Step2: Substitute $x=-8$

$m(-8)-12 = (-8)^2+2(-8)-4$

Step3: Simplify right-hand side

$-8m-12 = 64-16-4$
$-8m-12 = 44$

Step4: Solve for $m$

$-8m = 44+12$
$-8m = 56$
$m = \frac{56}{-8} = -7$

Question 10

Step1: Set functions equal at $x=c$

For continuity, $x^2-6 = 4x-10$ at $x=c$.

Step2: Rearrange to quadratic equation

$c^2-4x+4 = 0$

Step3: Factor the quadratic

$(c-2)^2 = 0$

Step4: Solve for $c$

$c = 2$

Answer:

$m = -7$
$c = 2$