Science:Math Exam Resources/Courses/MATH110/December 2014/Question 03 (e)/Solution 1

As we mentioned in the Hint, $g$ must be continuous at $x=4$.

This means $g(4)=\lim _{x\to 4^{+}}g(x)=\lim _{x\to 4^{-}}g(x)$.

Since we have $g(4)=\lim _{x\to 4^{+}}g(x)=\lim _{x\to 4^{+}}(mx+b)=4m+b$ and $\lim _{x\to 4^{-}}g(x)={\sqrt {4}}=2$,

for the continuity at $x=4$, we need $4m+b=2$.

On the other hand, the slope of $g$ also should be matched. i.e., $\lim _{x\to 4^{+}}g'(x)=\lim _{x\to 4^{-}}g'(x)$.

The left-hand side can be evaluated by

$\lim _{x\to 4^{+}}g'(x)=\lim _{x\to 4^{+}}(mx+b)'=m$,

while the right hand side becomes

$\lim _{x\to 4^{-}}g'(x)=\lim _{x\to 4^{-}}({\sqrt {x}})'=\lim _{x\to 4^{-}}{\frac {1}{2{\sqrt {x}}}}={\frac {1}{2{\sqrt {4}}}}={\frac {1}{4}}$.

Therefore, we get $m={\frac {1}{4}}$ and plugging this back to the equation $4m+b=2$, we obtain $b=1$.

Answer: $\color {blue}m={\frac {1}{4}},b=1$