SOLUTION:
Case: Local maxima and local minima
Given:
[tex]P(x)=5x^3+2x^2-42^{}[/tex]
Required: To find the local maxima and minima
Method:
[tex]\begin{gathered} P(x)=5x^3+2x^2-42^{} \\ \text{ } \end{gathered}[/tex]
First the find the first derivative:
[tex]\begin{gathered} P^{\prime}(x)=15x^2+4x \\ we\text{ make P'(x) = 0} \\ 15x^2+4x=\text{ 0} \\ x(15x+4)=0_{} \\ x=0\text{ or 15x+4 =0} \\ x=\text{ 0 or x= }\frac{-4}{15} \end{gathered}[/tex]
Let us take the points in the immediate neighbourhood of x = 0. The points are {-1, 1}.
[tex]\begin{gathered} \text{when x=-1} \\ P^{\prime}(x)=15x^2+4x \\ P^{\prime}(-1)=15(-1)^2+4(-1) \\ P^{\prime}(-1)=15(1)^2-4 \\ P^{\prime}(-1)=15^{}-4 \\ P^{\prime}(-1)=11 \\ \text{when x= 1} \\ P^{\prime}(x)=15x^2+4x \\ P^{\prime}(1)=15(1)^2+4(1) \\ P^{\prime}(1)=15^{}+4 \\ P^{\prime}(1)=19 \end{gathered}[/tex]
Let us take the points in the immediate neighbourhood of x = -4/15. The points are {-1, 0}
[tex]\begin{gathered} \text{when x= 0} \\ P^{\prime}(x)=15x^2+4x \\ P^{\prime}(0)=15(0)^2+4(0) \\ P^{\prime}(0)=0^{}+0 \\ P^{\prime}(0)=0^{} \end{gathered}[/tex]
Final answer:
The local minima is at x= 0 while the local maxima is at x= -4/15
