Cho hàm số y = f(x) có đạo hàm f'(x)= x^2 (x - 1)(13x - 15)^3. Khi đó số

* Hướng dẫn giải

Đáp án D

Ta có 

$$\begin{gathered} g\left(x\right)=f\left(\frac{5x}{x^2+4}\right)\Rightarrow g\text{'}\left(x\right)={\left(\frac{5x}{x^2+4}\right)}^{\text{'}}\text{'}f\text{'}\left(\frac{5x}{x^2+4}\right) \\ =\frac{5\left(4-x^2\right)}{{\left(x^2+4\right)}^2}.f\text{'}\left(\frac{5x}{x^2+4}\right) \end{gathered}$$

Mà 

$$\begin{gathered} f\text{'}\left(\frac{5x}{x^2+4}\right)=\frac{25x^2}{{\left(x^2+4\right)}^2}.\left(\frac{5x}{x^2+4}-1\right).{\left(\frac{65x}{x^2+4}-15\right)}^3 \\ =\frac{25x^2}{{\left(x^2+4\right)}^2}.\frac{\left(x-1\right)\left(x-4\right)}{x^2+4}.\frac{{\left(x-3\right)}^3{\left(15x-20\right)}^3}{{\left(x^2+4\right)}^3} \end{gathered}$$

Do đó $g\text{'}\left(x\right)=\frac{125x^2\left(4-x^2\right)\left(x-1\right)\left(x-4\right){\left(x-3\right)}^3{\left(15x-20\right)}^3}{{\left(x^2+4\right)}^8}$

Suy ra hàm số $y=g\left(x\right)$ có 6 điểm cực trị $x=\left\{\pm 2;1;4;3;\frac{4}{3}\right\}$