The function f(x) = x^3 - 48x has a local minimum at x = and a local maximum at x = ?
f'(x) = 3x^2 - 48
Set this equal to 0:
3x^2 - 48 = 0
Add 48 to each side:
3x^2 = 48
Divide each side by 3:
x^2 = 16
Therefore, x = -4, 4
Test f(4)
f(4) = 4^3 - 48(4)
f(4) = 64 - 192
f(4) = -128 <-- Local minimum
Test f(-4)
f(-4) = -4^3 - 48(-4)
f(-4) = -64 + 192
f(-4) = 128 <-- Local maximum
f'(x) = 3x^2 - 48
Set this equal to 0:
3x^2 - 48 = 0
Add 48 to each side:
3x^2 = 48
Divide each side by 3:
x^2 = 16
Therefore, x = -4, 4
Test f(4)
f(4) = 4^3 - 48(4)
f(4) = 64 - 192
f(4) = -128 <-- Local minimum
Test f(-4)
f(-4) = -4^3 - 48(-4)
f(-4) = -64 + 192
f(-4) = 128 <-- Local maximum