Question
The function $f(x)=x^5-5 x^4+5 x^3-1$ has
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Answer
$(d) :$ Given, $f(x)=x^5-5 x^4+5 x^3-1$
Then, $f^{\prime}(x)=5 x^4-20 x^3+15 x^2$
$\Rightarrow f^{\prime \prime}(x)=20 x^3-60 x^2+30 x$
Now, $f^{\prime}(x)=0 \Rightarrow 5 x^2\left(x^2-4 x+3\right)=0$
$\Rightarrow x=0,1,3$
$f^{\prime \prime}(1)=-10<0 \text { and } f^{\prime \prime}(3)=90>0$
$f^{\prime \prime}(0)=0 \text { and } f^{\prime \prime \prime}(0) \neq 0$
So, $x=0$ is a point of inflexion.
$\therefore f(x)$ has maximum at $x=1$ and minimum at $x=3$.
Then, $f^{\prime}(x)=5 x^4-20 x^3+15 x^2$
$\Rightarrow f^{\prime \prime}(x)=20 x^3-60 x^2+30 x$
Now, $f^{\prime}(x)=0 \Rightarrow 5 x^2\left(x^2-4 x+3\right)=0$
$\Rightarrow x=0,1,3$
$f^{\prime \prime}(1)=-10<0 \text { and } f^{\prime \prime}(3)=90>0$
$f^{\prime \prime}(0)=0 \text { and } f^{\prime \prime \prime}(0) \neq 0$
So, $x=0$ is a point of inflexion.
$\therefore f(x)$ has maximum at $x=1$ and minimum at $x=3$.
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