Question

Find the local maxima or local minima of $f\left(x\right)=x^3-6x^2+9x+15=0$ Also, find the local maximum or local minimum values as the case may be ?

Answer 1

Here,

$f\left(x\right)=x^3-6x^2+9x+15=0$

$\Rightarrow f^{\prime }\left(x\right)=3x^2-12x+9.$

For local maxima and minima we must have $f^{\prime }\left(x\right)=0$

Now,

$f^{\prime }\left(x\right)=0$ $\Rightarrow 3\left(x^2-4x+3\right)=0$

$\Rightarrow 3\left(x-3\right)\left(x-1\right)=0\Rightarrow x=3orx=1.$

Case. ! when $x=3$

In this case , when $x$ is slightly less than $3$ then $f^{\prime }\left(x\right)=3(x-3)9x-1)$ is negative and when $x$ is slightly more than $3$ then $f^{\prime }\left(x\right)$ is positive.

Thus, $f^{\prime }\left(x\right)$ changes from negative to positive as $x$ increases through $3$.

So, $x=3$ is a point of local minimum.

Case 2, when $x=1$

In this case, when $x$ is slightly less than $1$ then $f^{\prime }\left(x\right)=3(x-3)(x-1)$ is positive and when $x$ is slightly more than $1$ then $f^{\prime }\left(x\right)$ is negative.

Thus, $f^{\prime }\left(x\right)$ changes sign from positive to negative as $x$ increases through $1$.

So, $x=1$ is a point of local maximum.

Local maximum value $=f\left(1\right)=19$.