Question

The minimum value of $f\left(x\right)=-2x^2+5x+4\forall x\in [0,3]$ is __

• A. 1
• B. 1.0
• C. 1.00

Answer 1

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

$f\left(x\right)=-2x^2+5x+4$

$a<0\Rightarrow$ Graph of $f\left(x\right)$ will have maximum value at $x=\frac{-b}{2a}=\frac{5}{4}$

As $0<\frac{5}{4}<3$, minimum of $f\left(0\right)$ & $f\left(3\right)$ will be minimum value of $f\left(x\right)$ in $x\in [0,3]$.

So, $f\left(0\right)=4$

$f\left(3\right)=-18+15+4=1$

So, $f\left(3\right)=1$ is minimum value.