Question

The graph of $f\left(x\right)=\frac{x^5}{20}-\frac{x^4}{12}+5$ has

• A. no local extreme, one point of inflection
• B. two local maximum, one local minimum, two point of inflection
• C. one local maximum, one local minimum, one point of inflection
• D. one local maximum, one local minimum, two point of inflection

Answer 1

Answer: (C)

one local maximum, one local minimum, one point of inflection

$f\left(x\right)=\frac{x^5}{20}-\frac{x^4}{12}+5$
$f^{\prime }\left(x\right)=\frac{x^4}{4}-\frac{x^3}{3}=\frac{x^3}{12}(3x-4)$
For maxima or minima,
$f^{\prime }\left(x\right)=0$
$\Rightarrow x=0,\frac{4}{3}$
$f^{\prime }\left(x\right)>0$ for $x\in (-\infty ,0)\cup (\frac{4}{3},\infty )$

$f^{\prime }\left(x\right)<0$ for $x\in (0,\frac{4}{3})$

$f^{\prime \prime }\left(x\right)=0\Rightarrow x^3-x^2=0$ at $x=0$ and $1$.

$f^{\prime \prime \prime }\left(x\right)=0\Rightarrow 3x^2-2x=0$ at $x=0$ and $\frac{2}{3}$

$\therefore x=0$ cannot be a point of inflection.

Hence,

$f\left(x\right)$ have one point of local maxima, one point of local minima, and one point of inflection.