What is the local maximum value of the function? (Round answer to the nearest thousandth)

[tex]g(x)=3x^3+3x^2-30x+24[/tex]

Respuesta :

Answer:

72.578 to the nearest thousandth.

Step-by-step explanation:

g(x)=3x^3 + 3x^2 - 30x + 24

Find the derivative:

g'(x) = 9x^2 + 6x - 30.

This equals zero for values of turning points:

9x^2 + 6x -30 = 0

3(3x^2 + 2x - 10) = 0

x =  [-(2) +/- √(2^2 - 4*3*-10)] / 2*3

x = (-2 +/- √124) / 6

x =  (-2 + 11.136)/6,  (-2 - 11.136)/6

x = 1.523, -2.189.

To check which value of x gives a maximum value for f(x) use the second derivative test:

g"(x) =  18x + 6

when x = -2.189  g"(x) = 18(-2.189) + 2 which is negative so this value gives a local maximum.

Local maximum value for f(x)

= 3(-2.189)^3 + 3(-2.189)^2 - 30(-2.189) + 24

= 72.578 answer.