Given:
[tex]f\mleft(x\mright)=4x^3-5x^2[/tex]Find-: Local minimum and local maximum and inflection point.
Sol:
Derivative of function.
[tex]\begin{gathered} f\mleft(x\mright)=4x^3-5x^2 \\ f^{\prime}\left(x\right)=12x^2-10x \\ f^{\prime}\left(x\right)=2x\left(6x-5\right) \end{gathered}[/tex]The critical point is:
[tex]\begin{gathered} f^{\prime}\left(x\right)=0 \\ 2x\left(6x-5\right)=0 \\ 2x=0;6x-5=0 \\ x=0;x=\frac{5}{6} \end{gathered}[/tex]Local minima is:
[tex]\left(x,f\lparen x\right))=\lparen\frac{5}{6},-1.157)[/tex]Local minima at x=5/6
Local maxima at x=0
Inflection point.
[tex]\begin{gathered} f=4x^3-5x^2 \\ \text{ Inflection point} \\ x=\frac{5}{12} \end{gathered}[/tex]