Question

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

• A. one local maximum and one local minimum
• B. one local minimum and one point of inflection.
• C. one local maximum, one local minimum and one point of inflection.
• D. one local maximum, one local minimum and two points of inflection.

Answer 1

The correct option is C one local maximum, one local minimum and one point of inflection.

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

$f^{\prime }\left(x\right)=\frac{x^4}{4}-\frac{x^3}{3}=\frac{x^3}{12}(3x-4)$
Now, $f^{\prime }\left(x\right)=0$ gives $x=0,\frac{4}{3}$
At $x=0^{-},f^{\prime }\left(x\right)>0$
At $x=0^{+},f^{\prime }\left(x\right)<0$
So, $f^{\prime }\left(x\right)$ change its sign from $+ve$ to $-ve$ at $x=0.$
Hence, at $x=0$, we have maxima.

At $x={\left(\frac{4}{3}\right)}^{-},f^{\prime }\left(x\right)<0$
At $x={\left(\frac{4}{3}\right)}^{+},f^{\prime }\left(x\right)>0$
So, $f^{\prime }\left(x\right)$ change its sign from $-ve$ to $+ve$ at $x=\frac{4}{3}.$
Hence, at $x=\frac{4}{3}$, we have minima.

Now, $f^{\prime \prime }\left(x\right)=x^3-x^2=x^2(x-1)$
$f^{\prime \prime }\left(x\right)=0$ gives $x=0,1$
At $x=1^{-},f^{\prime }\left(x\right)<0$
At $x=1^{+},f^{\prime }\left(x\right)>0$
So, $f^{\prime }\left(x\right)$ change its sign at $x=1.$
Hence, at $x=1$, is the point of inflection.