For all $a,b\in R$ the function $f(x)=3x^4-4x^3+6x^2+ax+b$ has :
$(A)$no extrema
$(B)$exactly one extremum
(C)exactly two extrema
(D)three extrema
$f(x)=3x^4-4x^3+6x^2+ax+b\Rightarrow f'(x)=12x^3-12x^2+12x+a$
For extremum,$f'(x)=12x^3-12x^2+12x+a=0$
What should i do now to determine the number of extrema? It appears that there are 3 extrema,because $f'(x)=0$ is a cubic equation. But in the answer in the book,exactly one extremum is the answer. Can someone please explain me why is it so?
The second derivative of the function is given by
$f''(x)=12(3x^2-2x+1)$
Which is positive for all x∈R, regardless of $a,b$ meaning there are no inflection points, and that there is at most one minimum to $f(x)$. Also: $f'(-\infty)<0$ and $f'(\infty)>0$. Meaning there is atleast one minimum to the function.
From this you can conclude the function has one extremum.