What is the extreme f(x)=7x^4-6x^2-1 ?

The extreme f(x)=7x^4-6x^2-1 is Minimum(-sqrt(3/7),-16/7),Maximum(0,-1),Minimum(sqrt(3/7),-16/7)

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extreme f(x)=7x^4-6x^2-1

extreme points

f (x) = 7 x^{4} - 6 x^{2} - 1

Minimum (- \frac{3}{7}, - \frac{16}{7}), Maximum (0, - 1), Minimum (\frac{3}{7}, - \frac{16}{7})

Solution steps

f^{'} (x) = 28 x^{3} - 12 x

Find intervals:

Decreasing

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

Increasing

: - \frac{3}{7} < x < 0,

Decreasing

: 0 < x < \frac{3}{7},

Increasing

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

Plug

x = - \frac{3}{7}

into

7 x^{4} - 6 x^{2} - 1 : - \frac{16}{7}

Minimum (- \frac{3}{7}, - \frac{16}{7})

Plug

x = 0

into

7 x^{4} - 6 x^{2} - 1 : - 1

Maximum (0, - 1)

Plug

x = \frac{3}{7}

into

7 x^{4} - 6 x^{2} - 1 : - \frac{16}{7}

Minimum (\frac{3}{7}, - \frac{16}{7})

Minimum (- \frac{3}{7}, - \frac{16}{7}), Maximum (0, - 1), Minimum (\frac{3}{7}, - \frac{16}{7})

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

f (x) = 4 x + \frac{4}{x}

f (x) = 7 x^{4} - 6 x^{2} - 6

f (x) = \frac{x ^{2} + x + 1}{x ^{2} - x + 1}

f (x) = 6 - 8 x, - 8 \leq x \leq 6

What is the extreme f(x)=7x^4-6x^2-1 ?
The extreme f(x)=7x^4-6x^2-1 is Minimum(-sqrt(3/7),-16/7),Maximum(0,-1),Minimum(sqrt(3/7),-16/7)