Question

f(x) = x³ $-$ 6x² + 9x + 15

Answer 1

$$\begin{gathered} \mathrm{Given}:f\left(x\right)=x^3-6x^2+9x+15 \\ \Rightarrow f\text{'}\left(x\right)=3x^2-12x+9 \\ \mathrm{For}\mathrm{a}\mathrm{local}\mathrm{maximum}\mathrm{or}\mathrm{a}\mathrm{local}\mathrm{minimum},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)=0 \\ \Rightarrow 3x^2-12x+9=0 \\ \Rightarrow x^2-4x+3=0 \\ \Rightarrow \left(x-1\right)\left(x-3\right)=0 \\ \Rightarrow x=1\mathrm{or}3 \end{gathered}$$



Since f '(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
${\left(3\right)}^3-6{\left(3\right)}^2+9\left(3\right)+15=27-54+27+15=15$

Since f '(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
${\left(1\right)}^3-6{\left(1\right)}^2+9\left(1\right)+15=1-6+9+15=19$