Question

The polynomial P(x) = 5x^3 + 2x^2- 42 has ___ local maxima and minima

Answer 1

Case: Local maxima and local minima

Given:

`P(x)=5x^3+2x^2-42^{}`

Required: To find the local maxima and minima

Method:

`\begin{gathered} P(x)=5x^3+2x^2-42^{} \\ \text{ } \end{gathered}`

First the find the first derivative:

`\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= }(-4)/(15) \end{gathered}`

Let us take the points in the immediate neighbourhood of x = 0. The points are {-1, 1}.

`\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}`

Let us take the points in the immediate neighbourhood of x = -4/15. The points are {-1, 0}

`\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}`

Final answer:

The local minima is at x= 0 while the local maxima is at x= -4/15