What is the extreme f(x)=12x^2-48x+20[0.3] ?

The extreme f(x)=12x^2-48x+20[0.3] is Minimum(2,-42)

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extreme f(x)=12x^2-48x+20[0.3]

extreme points

f (x) = 12 x^{2} - 48 x + 20 [0.3]

Minimum (2, - 42)

Solution steps

f^{'} (x) = 24 x - 48

Find intervals:

Decreasing

: - \infty < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 2

into

12 x^{2} - 48 x + 20 [0.3] : - 42

Minimum (2, - 42)

f (x) = \frac{3}{x - 4}

y = x^{2} + 4 x - x

f (x) = 4 cos (2 x) + 1

0.5 x^{3} - 3 x^{2} + 4.5 x

- 2 x^{3} - 3 x^{2} + 12 x

What is the extreme f(x)=12x^2-48x+20[0.3] ?
The extreme f(x)=12x^2-48x+20[0.3] is Minimum(2,-42)