Question

If $f\left(x\right)=\{\begin{array}{c}x-5;x\le 1\\ 4x^2-9;1<x\le 2\\ 3x^2+4;x>2\end{array}$
Then $f\left(2^{+}\right)-f\left(2^{-}\right)=$

• A. $0$
• B. $2$
• C. $9$
• D. $4$

Answer 1

Answer: (C)

$9$

Given: $f\left(x\right)=\{\begin{array}{c}x-5;x\le 1\\ 4x^2-9;1<x\le 2\\ 3x^2+4;x>2\end{array}$

$f\left(2^{+}\right)=\underset{x\to 2^{+}}{lim}f\left(x\right)$

$=\underset{h\to 0}{lim}3(2+h{)}^2+4$

$f\left(2^{+}\right)=16$

$f\left(2^{-}\right)=\underset{x\to 2^{-}}{lim}f\left(x\right)$

$=\underset{h\to 0}{lim}4(2-h{)}^2-9$

$f\left(2^{-}\right)=7$

So, $f\left(2^{+}\right)-f\left(2^{-}\right)=16-7=9$