Question

If $f\left(x\right)=(\begin{array}{c}3x^2+12x-1,-1\le x\le 2\\ 37-x,2<x\le 3\end{array}$, then

• A. f(x) is increasing in [-1, 2]
• B. f(x) is continuous in [-1, 3]
• C. f '(2) does not exist
• D. f(x) has a maximum value at x = 2

Answer 1

Answer: (A), (B), (C), (D)

f(x) has a maximum value at x = 2

We have,
$f\left(x\right)=(\begin{array}{c}3x^2+12x-1,-1\le x\le 2\\ 37-x,2<x\le 3\end{array}$
Then, in [-1, 2],
$f^{\prime }\left(x\right)=6x+12.f^{\prime }\left(x\right)=0\Rightarrow x=-2\text{Thus, f(x) decreases in}(-\infty ,-2)\text{and increases in}(-2,\infty )$
Also, $f\left(2^{-}\right)=3(2{)}^2+12(2)-1=35andf(2^{+})=37-2=35$
Hence, f(x) is continuous.
$f\text{'}\left(x\right)=(\begin{array}{c}6x+12,-1\le x\le 2\\ -1,2<x\le 3\end{array}\therefore f\text{'}(2^{-})=24\mathrm{and}f\text{'}(2^{+})=-1\mathrm{Hence},f(x)\mathrm{is}\mathrm{non}-\mathrm{differentiable}\mathrm{at}x=2.\mathrm{Also},f\text{'}(2^{+})<f(2)\mathrm{and}f\text{'}(2^{-})<f(2).\mathrm{Hence},x=2\mathrm{is}\mathrm{the}\mathrm{point}\mathrm{of}\mathrm{maxima}.$