What is the extreme f(x)=4x^2+3x^2-6x+1 ?

The extreme f(x)=4x^2+3x^2-6x+1 is Minimum(3/7 ,-2/7)

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extreme f(x)=4x^2+3x^2-6x+1

extreme points

f (x) = 4 x^{2} + 3 x^{2} - 6 x + 1

Minimum (\frac{3}{7}, - \frac{2}{7})

Solution steps

f^{'} (x) = 14 x - 6

Find intervals:

Decreasing

: - \infty < x < \frac{3}{7},

Increasing

: \frac{3}{7} < x < \infty

Plug

x = \frac{3}{7}

into

4 x^{2} + 3 x^{2} - 6 x + 1 : - \frac{2}{7}

Minimum (\frac{3}{7}, - \frac{2}{7})

\frac{4 x + 13}{- 2 x - 6}

y = x^{4} - 12 x^{3} + 48 x^{2} - 64 x

f (x, y) = x^{2} + x y + 2 x + 5 y - 1

0.01 x^{3} - 0.45 x^{2} + 2 x + 300

x e^{\frac{2}{x}}

What is the extreme f(x)=4x^2+3x^2-6x+1 ?
The extreme f(x)=4x^2+3x^2-6x+1 is Minimum(3/7 ,-2/7)