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Popular >

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

Solution

extreme points

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

Solution

Minimum (\frac{1 - 3}{2}, \frac{( 1 - 3 ) ^{4}}{16} + 3 - 1), Maximum (\frac{1}{2}, \frac{21}{16}), Minimum (\frac{1 + 3}{2}, \frac{( 1 + 3 ) ^{4}}{16} - 1 - 3)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < \frac{1 - 3}{2},

Increasing

: \frac{1 - 3}{2} < x < \frac{1}{2},

Decreasing

: \frac{1}{2} < x < \frac{1 + 3}{2},

Increasing

: \frac{1 + 3}{2} < x < \infty

Plug

x = \frac{1 - 3}{2}

into

x^{4} - 2 x^{3} + x + 1 : \frac{( 1 - 3 ) ^{4}}{16} + 3 - 1

Minimum (\frac{1 - 3}{2}, \frac{( 1 - 3 ) ^{4}}{16} + 3 - 1)

Plug

x = \frac{1}{2}

into

x^{4} - 2 x^{3} + x + 1 : \frac{21}{16}

Maximum (\frac{1}{2}, \frac{21}{16})

Plug

x = \frac{1 + 3}{2}

into

x^{4} - 2 x^{3} + x + 1 : \frac{( 1 + 3 ) ^{4}}{16} - 1 - 3

Minimum (\frac{1 + 3}{2}, \frac{( 1 + 3 ) ^{4}}{16} - 1 - 3)

Minimum (\frac{1 - 3}{2}, \frac{( 1 - 3 ) ^{4}}{16} + 3 - 1), Maximum (\frac{1}{2}, \frac{21}{16}), Minimum (\frac{1 + 3}{2}, \frac{( 1 + 3 ) ^{4}}{16} - 1 - 3)

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Frequently Asked Questions (FAQ)

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