solve this

Given : $f(x)=x^{3}-6 x^{2}+9 x+15$

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

For a local maximum or a local minimum, we must have

$f^{\prime}(x)=0$

$\Rightarrow 3 x^{2}-12 x+9=0$

$\Rightarrow x^{2}-4 x+3=0$

$\Rightarrow(x-1)(x-3)=0$

$\Rightarrow x=1$ or 3

Since $f^{\prime}(x)$ changes from negative to positive when $x$ increases through $3, x=3$ is the point of local minima. The local minimum value of $f(x)$ at $x=3$ is given by

$(3)^{3}-6(3)^{2}+9(3)+15=27-54+27+15=15$

Since $f^{\prime}(x)$ changes from positive to negative when $x$ increases through $1, x=1$ is the point of local maxima. The local maximum value of $f(x)$ at $x=1$ is given by

$(1)^{3}-6(1)^{2}+9(1)+15=1-6+9+15=19$