Question

hw 5 - continuity section 1.4: problem 10 (1 point) find a value of the constant k, if possible, at which f(x) = {kx^2 x <= -7; 7x + k x > -7} is continuous everywhere. k = (enter
one\ if no such value exists). write your answer as a simplified fraction.

Explanation:

Step1: Recall continuity condition

For a function to be continuous at $x = - 7$, $\lim_{x
ightarrow - 7^{-}}f(x)=\lim_{x
ightarrow - 7^{+}}f(x)$.

Step2: Calculate left - hand limit

$\lim_{x
ightarrow - 7^{-}}f(x)=\lim_{x
ightarrow - 7^{-}}kx^{2}=k(-7)^{2}=49k$.

Step3: Calculate right - hand limit

$\lim_{x
ightarrow - 7^{+}}f(x)=\lim_{x
ightarrow - 7^{+}}(7x + k)=7(-7)+k=-49 + k$.

Step4: Set left - hand and right - hand limits equal

Set $49k=-49 + k$.

Step5: Solve for k

Subtract $k$ from both sides: $49k - k=-49$, so $48k=-49$. Then $k =-\frac{49}{48}$.

Answer:

$-\frac{49}{48}$