Question

Cho hàm số \(f(x) = \sqrt {3x + 1} \). Đặt \(g(x) = f(1) + 4\left( {{x^2} - 1} \right)f'(1)\). Tính \(g(2)\).

Answer 1

Ta có: \(f'\left(x\right)=\dfrac{3}{2\sqrt{3x+1}}\)

Do đó, \(f\left(1\right)=2,f'\left(1\right)=\dfrac{3}{4}\)

Vậy \(g\left(2\right)=f\left(1\right)+4\left(2^2-1\right)f'\left(1\right)=2+12\cdot\dfrac{3}{4}=11\)