Question

For certain curves $y=f\left(x\right);x\in [0,2]$ satisfying $\frac{d^{2}y}{d{x}^2}=6x-4,f\left(x\right)$ has local minimum value $5$ when $x=1$, then

• A. Number of critical point of the curve is $2$
• B. Global minimum value of $f\left(x\right)$ is $5$
• C. Global maximum value of $f\left(x\right)$ is $7$
• D. $f\left(0\right)=5$

Answer 1

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

$f\left(0\right)=5$

$\frac{d^{2}y}{d{x}^2}=6x-4\text{Integrating w.r.t.}x\frac{dy}{dx}=3x^2-4x+c$
When $x=1,\frac{dy}{dx}=0$
$0=3-4+c\Rightarrow c=1\Rightarrow \frac{dy}{dx}=3x^2-4x+1\Rightarrow y=x^3-2x^2+x+c_{1}x=1,y=5\Rightarrow c_1=5\Rightarrow y=f\left(x\right)=x^3-2x^2+x+5$
For critical points,
$\frac{dy}{dx}=0\Rightarrow 3x^2-4x+1=0\Rightarrow (3x-1)(x-1)=0\Rightarrow x=1,\frac{1}{3}$
When $x=\frac{1}{3}$
$\frac{d^{2}y}{d{x}^2}<0$
Therefore, $x=\frac{1}{3}$ is point of local maximum and
$x=1$ is point of local minimum.
Now,
$f\left(1\right)=5f\left(\frac{1}{3}\right)=\frac{139}{27}f\left(0\right)=5f\left(2\right)=7$

Hence,
Number of critical points is $2$,
Global maximum is $7$
Global minimum is $5$.