Question

find the extreme value of f(x)=3x^2+x-1​

Answer 1

Answer:

x=0

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Step-by-step explanation:

Answer 2

Answer:

Differentiate with respect to x

f'(x) = 6x + 1

For max and min

f'(x) = 0

6x + 1 = 0

x = -1/6

Differentiate [1] with respect to x

f''(x) = 6

f(x) is minimum at x = -1/6

and minimum value is

$f( \frac{ - 1}{6} ) = 3 {( \frac{ - 1}{6} )}^{2} - \frac{1}{6} + 1 \\ = \frac{1}{12} - \frac{1}{6} + 1 \\ = \frac{11}{12}$