The interval in which the function, f(x) = 7 – 4x – x2 is strictly increasing is
Answer & explanation
Correct answer: option 3
$f(x) = 7 - 4x - x^2$
$f'(x) > 0$
$f'(x)=\frac{d}{dx}(7 - 4x - x^2)$
{$\frac{d}{dx}x^n=nx^{n-1}$} $\frac{d}{dx}(c)=0$
$= 0-4-2x⇒-(4+2x)$
$-(4+2x)>0$
$4+2x<0⇒2x<-4$
$x<-2$
$x∈(-∞, -2) $