Find the maximum and the minimum values (if any) of the function $f(x) = 16x^2 - 16x+28$.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Maximum value: None; Minimum value: 24
Given $f(x) = 16x^2 - 16x + 28 = 16\left(x^2-x+\frac{7}{4}\right)$
$= 16\left[\left(x^2 - x + \frac{1}{4}\right) + \frac{7}{4} -\frac{1}{4}\right] = 16\left[\left(x − \frac{1}{2}\right)^2 + \frac{3}{2}\right]$
$≥16\left(0+\frac{3}{2}\right)$, for all $x ∈ R$ $(∵(x-\frac{1}{2})^2≥0$, for all $x∈ R)$
$⇒ f(x) > 24$ for all $x ∈ R$.
Therefore, the minimum value of f(x) is 24, it has no maximum value.