Can anyone find the absolute minimum value of the function f(x) = (1/4)x^4 + x^3 - 5x^2?

Answer 1

f(x) = (1/4)x^4 + x^3 - 5x^2
f'(x) = x^3 +3x^2 - 10x
0 = x^3 +3x^2 - 10x
0 = x(x+5)(x-2)
x = 0, -5, +2
Try those
f(0) = 0
f (2) = 1/4(16) + 8 -20 = -8
f(-5) = 1/4(625) -125 - 125 = -93.75 (absolute minimum)

Answer 2

f(x) = (1/4)x^4 + x^3 -5x^2
f'(x) = x^3 + 3x^2 - 10x
f'(x) = x(x + 5)(x-2)

extrema are at x = -5, x = 0 and x = 2

f(-5) = -93.75
f(0) = 0
f(2) = -8

Since f(x) is positive for values x < -5 and x > 2 the absolute minimum of f(x) is at x = -5

Answer 3

f' = x³ + 3x² - 10x
f' = 0
=> x³ + 3x² - 10x = 0
=> x(x+5)(x-2) = 0
=> x = 0,-5,2

for x=0, f(x) = 0
for x=2, f(x) = -8
for x = -5, f(x) = 625/4 - 125 - 125
= -93.75

=> the absolute minimum value of f(x) is -93.75

Answer 4

f' = x^3 + 3x^2 - 10x

x(x^2 + 3x - 10) = 0
x(x+5)(x-2)=0
x=-5,0,2

Using a number line:
f' (-).....(+)...(-)....(+)
<--- -5 --- 0 --- 2 ---->

So there is a local min at x=-5 and x=2
and a local max at x=0

Since the function is defined on all real numbers (it is a polynomial), there are no bounds to check.

Therefore all we need to check are the two local mins to find the absolute min.
f(-5) = -93.75
f(2) = -8

Therefore the absolute min is located at x=-5, and the absolute minimum value is -93.75.