Question

The function $x^5-5x^4+5x^3-10$ has a maximum, when $x=$

• A. $3$
• B. $2$
• C. $1$
• D. $0$

Answer 1

The correct option is A $3$

To obtain the absolute maxima or minima for the function $f\left(x\right)$

Find $f^{\prime }\left(x\right)$ and put $f^{\prime }\left(x\right)=0$

Given

$x^5-5x^4+5x^3-10$

$f\left(x\right)=x^5-5x^4+5x^3-10$

$f\left(x\right)=x^5-5x^4+5x^3-10$

$\frac{df\left(x\right)}{dx}=f^{\prime }\left(x\right)=5x^4-20x^3+15x^2$

$5x^4-20x^3+15x^2=0$

$5x^2(x-3)(x-1)=0$

$x=0,x=1,x=3$

To find the maximum find $f^{\prime \prime }\left(x\right)$ and substitute the value of $x$

$\frac{d{f}^{\prime }\left(x\right)}{dx}=f^{\prime \prime }\left(x\right)=20x^3-60x^2+30x$

When, $x=3$

$f^{\prime \prime }\left(x\right)=20x^3-60x^2+30x=20(3{)}^3-60(3{)}^2+30\left(3\right)=90>0$

When, $x=0$

$f^{\prime \prime }\left(x\right)=20x^3-60x^2+30x=20(0{)}^3-60(0{)}^2+30\left(0\right)=0$

When, $x=1$

$f^{\prime \prime }\left(x\right)=20x^3-60x^2+30x=20(1{)}^3-60(1{)}^2+30\left(1\right)=-10<0$

$\therefore$ The given function has the maximum value when $x=3$