1) To find the relative maxima of a function, we need to perform the first derivative test. It tells us whether the function has a local maximum, minimum r neither.
[tex]\begin{gathered} f^{\prime}(x)=\frac{d}{dx}\mleft(8x^3-3x^2\mright) \\ f^{\prime}(x)=\frac{d}{dx}\mleft(8x^3\mright)-\frac{d}{dx}\mleft(3x^2\mright) \\ f^{\prime}(x)=24x^2-6x \end{gathered}[/tex]2) Let's find the points equating the first derivative to zero and solving it for x:
[tex]\begin{gathered} 24x^2-6x=0 \\ x_{}=\frac{-\left(-6\right)\pm\:6}{2\cdot\:24},\Rightarrow x_1=\frac{1}{4},x_2=0 \\ f^{\prime}(x)>0 \\ 24x^2-6x>0 \\ \frac{24x^2}{6}-\frac{6x}{6}>\frac{0}{6} \\ 4x^2-x>0 \\ x\mleft(4x-1\mright)>0 \\ x<0\quad \mathrm{or}\quad \: x>\frac{1}{4} \\ f^{\prime}(x)<0 \\ 24x^2-6x<0 \\ 4x^2-x<0 \\ x\mleft(4x-1\mright)<0 \\ 0Now, we can write out the intervals, and combine them with the domain of this function since it is a polynomial one that has no discontinuities:[tex]\mathrm{Increasing}\colon-\infty\: 3) Finally, we need to plug the x-values we've just found into the original function to get their corresponding y-values:[tex]\begin{gathered} f(x)=8x^3-3x^2 \\ f(0)=8(0)^3-3(0)^2 \\ f(0)=0 \\ \mathrm{Maximum}\mleft(0,0\mright) \\ x=\frac{1}{4} \\ f(\frac{1}{4})=8\mleft(\frac{1}{4}\mright)^3-3\mleft(\frac{1}{4}\mright)^2 \\ \mathrm{Minimum}\mleft(\frac{1}{4},-\frac{1}{16}\mright) \end{gathered}[/tex]4) Finally, for the inflection points. We need to perform the 2nd derivative test:
[tex]\begin{gathered} f^{\doubleprime}(x)=\frac{d^2}{dx^2}\mleft(8x^3-3x^2\mright) \\ f\: ^{\prime\prime}\mleft(x\mright)=\frac{d}{dx}\mleft(24x^2-6x\mright) \\ f\: ^{\prime\prime}(x)=48x-6 \\ 48x-6=0 \\ 48x=6 \\ x=\frac{6}{48}=\frac{1}{8} \end{gathered}[/tex]Now, let's plug this x value into the original function to get the y-corresponding value:
[tex]\begin{gathered} f(x)=8x^3-3x^2 \\ f(\frac{1}{8})=8(\frac{1}{8})^3-3(\frac{1}{8})^2 \\ f(\frac{1}{8})=-\frac{1}{32} \\ Inflection\: Point\colon(\frac{1}{8},-\frac{1}{32}) \end{gathered}[/tex]