Question

If $f\left(x\right)=3x^2+12x-1$, $xϵ[-1,2]$
$37-x$, $xϵ[2,3]$

• A. $f\left(x\right)$ is monotonic increasing in $[-1,2]$
• B. $f\left(x\right)$ is continuous in $[-1,3]$
• C. $f\left(x\right)$ has the maximum value at $x=2$
• D. $f\left(x\right)$ does not exist

Answer 1

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

The correct options are
A $f\left(x\right)$ is monotonic increasing in $[-1,2]$
B $f\left(x\right)$ is continuous in $[-1,3]$
C $f\left(x\right)$ has the maximum value at $x=2$
D $f\left(x\right)$ does not exist

For $-1\le x\le 2$ , we have
$f\left(x\right)={3x}^2+12x-1f^{\prime }\left(x\right)=6x+12\ge 0\forall -1\le x\le 2$
Hence $f\left(x\right)$ is increasing in $[-1,2]$
Again function is algebraic polynomial therefore, it is continuous at $xϵ(-1,2)$ and $(2,3)$
and for continuity at $x=2$
${lim}_{x\to 2-}f\left(x\right)={lim}_{x\to 2-}({3x}^2+12x-1)={lim}_{h\to 0}[3{(2-h)}^2+12(2-h)-1]={lim}_{h\to 0}[3(4+h^2-4h)+24-12h-1]={lim}_{h\to 0}(12+{3h}^2-12h+24-12h-1)={lim}_{h\to 0}({3h}^2-24h+35)=35{lim}_{x\to 0}f\left(x\right)={lim}_{x\to 0}(37-x)={lim}_{h\to 2+}[37-(2+h)]=35f\left(2\right)=3.2^2+12.2-1=12+24-1=35$
Therefore $LHL=RHL=f\left(2\right)\Rightarrow$ function is continuous at
$x=2\Rightarrow$ function is continuous in $-1\le x\le 3$
Now $R{f}^{\prime }\left(2\right)=\underset{x\to 2+}{lim}\frac{f\left(x\right)-f\left(2\right)}{x-2}=\underset{h\to 0}{lim}\frac{f(2+h)-f\left(2\right)}{h}$
$=\underset{h\to 0}{lim}\frac{37-(2+h)-3(3\times 2^2+12\times 2-1)}{h}=\underset{h\to 0}{lim}\frac{-h}{h}=-1$
$L{f}^{\prime }\left(x\right)=\underset{x\to 2-}{lim}\frac{f\left(x\right)-f\left(2\right)}{x-2}=\underset{h\to 0}{lim}\frac{f(2-h)-f\left(2\right)}{-h}$
$=\underset{h\to 0}{lim}\frac{[[3{(2-h)}^2+12(2-h)-1]3\times 2^2+12\times 2-1]}{-h}$
$=\underset{h\to 0}{lim}\frac{[12+{3h}^2-12h+24=12h-1]-35}{-h}$
$=\underset{h\to 0}{lim}\frac{{3h}^2-24h+35-35}{-h}=\underset{h\to 0}{lim}\frac{3h-24}{-1}=24$
Since $R{f}^{\prime }\left(2\right)\ne f^{\prime }\left(2\right),f^{\prime }\left(2\right)$ does not exists
Again $f\left(x\right)$ increasing on $[-1,2]$ and is decreasing on $(2,3)$
it shows that $f\left(x\right)$ has maximum value at $x=2$