Question

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

Answer 1

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

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

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

So, f(0) = 4

f(3) = -18 + 15 + 4 = 1

so, f(3) = 1 is minimum value