Question

find the critical point of f(x) =4x^3+3x^2-6x+1​

Answer 1

Answer:

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

f'(x)=12x^2+6x-6

f''(x)=24x+6

f'''(x)=24≠0

f'(x)=0

12x^2+6x-6=0

12x^2+12x-6x-6

12x(x+1)-6(x+1)

(12x-6)(x+1)

12x-6=0 Either x+1=0

12x=6 x=-1

x=½

This is the required answer you need.

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