What is the extreme f(x)=(12*x^2-60x+44) ?

The extreme f(x)=(12*x^2-60x+44) is Minimum(5/2 ,-31)

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extreme f(x)=(12*x^2-60x+44)

extreme points

f (x) = (12 \cdot x^{2} - 60 x + 44)

Minimum (\frac{5}{2}, - 31)

Solution steps

f^{'} (x) = 24 x - 60

Find intervals:

Decreasing

: - \infty < x < \frac{5}{2},

Increasing

: \frac{5}{2} < x < \infty

Plug

x = \frac{5}{2}

into

12 \cdot x^{2} - 60 x + 44 : - 31

Minimum (\frac{5}{2}, - 31)

f (x) = x^{4} - 29 x^{2} + 100

f (x) = \frac{e ^{x}}{1 + 5 e ^{x}}

2 x + y - 3 = 0

f (x) = 2 x - 3

f (x) = 5 + x^{\frac{2}{5}}

What is the extreme f(x)=(12*x^2-60x+44) ?
The extreme f(x)=(12*x^2-60x+44) is Minimum(5/2 ,-31)