Given, $f(x)=x^{3}-6 x^{2}+9 x+15$
Differentiate with respect to $x$, we get, $f^{\prime}(x)=3 x^{2}-12 x+9=3\left(x^{2}-4 x+3\right)$ $=3(x-3)(x-1)$
For all maxima and minima,
$
\begin{array}{l}
f^{\prime}(x)=0 \\
=3(x-3)(x-1)=0 \\
=x=3,1
\end{array}
$
At $x=1, f^{\prime}(x)$ changes from positive to negative
Since, $x=1$ is a point of Maxima
At $x=3, f^{\prime}(x)$ changes from negative to positive
Since, $x=3$ is point of Minima.
Hence, local maxima value $f(1)=(1)^{3}-6(1)^{2}+9(1)+15=19$
Local minima value $f(3)=(3)^{3}-6(3)^{2}+9(3)+15=15$
Find the points of local maxima or local minima, if any, of the following functions, using the first derivative test. Also, find the local maximum or local minimum values, as the case may be: f (x) = x^3 – 6x^2 + 9x +15
Find the points of local maxima or local minima, if any, of the following functions, using the first derivative test. Also, find the local maximum or local minimum values, as the case may be: f (x) = x^3 – 6x^2 + 9x +15