Practicing Success
Practicing Success
CUET Preparation Today
The image of interval [–1, 3] under the mapping f : R → R given by $f (x) =4x^3- 12x$ is |
[8, 72] [–8, 72] [0, 8] none of these |
[–8, 72] |
$f (x) =4x^3- 12x$ $f'(x)=12x^2-12≥0$ for $x∈[-1,3]$ so $f_{min}=f(-1)=8,\,f_{max}=f(3)=72$ so for $x∈[-1,3]⇒f(x)∈[-8,72]$ |
