what are the max and min points for 2^3+4x^2-3x-18?

Answer 1

assuming the function is: 2x^3+4x^2-3x-18 and you're meant to figure out using calculus?


Max and min points can occur occur when the derivative is 0 or undefined. Find those points and figure out whether they are a min, a max, or neither.

f(x) = 2x^3+4x^2-3x-18
f'(x) = 6x^2 +8x -3

the derivative is a quadratic and therefore always defined, so you only need to find where it's 0 to obtain candidates for maxes and mins

6x^2+8x-3 = 0

Solve using quadratic formula

(-2/3) +/- sqrt(34)/6

~.30514 and -1.6384

look at the second derivative at these points

f''(x) = 12x+8

12x+8 = 0
x = -2/3

the second derivative is positive (concave up) on (-2/3, +inf) and negative on (-inf,-2/3).

so the function looks sort of

-1.6384 lies in the concave down region making it a local max
.30514 lies on the concave up portion making it a local min

Answer 2

2 and -5/4