Question

For all $a,b\in R$ the function $f\left(x\right)=3x^4-4x^3+6x^2+ax+b$ has-

• A. no extrenum
• B. exactly one extremum
• C. exactly two extremum
• D. three extremum

Answer 1

The correct option is B exactly one extremum

$f^{\prime }\left(x\right)=12x^3-12x^2+12x+a$
And
$f^{\prime \prime }\left(x\right)=36x^2-24x+12$
$=12(3x^2-2x+1)$
Now
$D=2^2-4\left(3\right)$
$=4-12$
$=-8$
Hence
$D<0$, thus
$f"\left(x\right)>0$ for all $x.$
Hence f(x) will only have minima, and no maxima.
Also $f"\left(x\right)$ has no real roots.
Therefore $f^{\prime }\left(x\right)$, $x$ will have atmost one real root.
Hence $f^{\prime }\left(x\right)=0$ will give us at most $1$ critical point.
Thus the conclusions are.
a) The above polynomial function has only one extremum, since it has atmost one critical point.
b) It attains a minimum only. It does not attain a maximum.