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

The extreme f(x)=3x^4+x-1 is Minimum(-1/(\sqrt[3]{12)},-(3^{2/3})/(4\sqrt[3]{4)}-1)

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

extreme points

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

Solution steps

f^{'} (x) = 12 x^{3} + 1

Find intervals:

Decreasing

Increasing

Plug

into

x + \frac{108}{x ^{2}}

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

f (x) = 5 x - ln (x)

f (x) = x^{4} - 12 x^{3} + 16 x^{2} + 5

f (x) = 4 x^{3} - 2 y^{2} - 12 x + 8 y

What is the extreme f(x)=3x^4+x-1 ?
The extreme f(x)=3x^4+x-1 is Minimum(-1/(\sqrt[3]{12)},-(3^{2/3})/(4\sqrt[3]{4)}-1)