Question

Global minimum value of $y=f\left(x\right)$ for $x\in [0,2]$ is

• A. $5$
• B. $7$
• C. $8$
• D. $9$

Answer 1

Answer: (A)

$5$

Integrating $\frac{d^{2}y}{d{x}^2}=6x-4,$ we get $\frac{dy}{dx}=3x^2-4x+A$
When $x=1,\frac{dy}{dx}=0$ so that $A=1$. Hence
$\frac{dy}{dx}=3x^2-4x+1$ ____(1)
Integrating, we get $y=x^3-2x^2+x+B$.
When $x=1,y=5$, so that $B=5$
Thus, we have $y=x^3-2x^2+x+5$
From equation (1), we get the critical points $x=1/3,x=1$
At the critical point $x=\frac{1}{3},\frac{d^{2}y}{d{x}^2}$ is -ve
Therefore, at $x=1/3$, y has a local maximum.
At $x=1,\frac{d^{2}y}{d{x}^2}$ is +ve
Therefore, at $x=1$ y has a local minimum.
Also $f\left(1\right)=5,f\left(\frac{1}{3}\right)=\frac{139}{27},f\left(0\right)=5,f\left(2\right)=7$
Hence the global maximum value $=7$
and the global minimum value $=5$