Question

Cho hàm số \(y=f\left(x\right)=4\sqrt{2x-6}\)

a) Dùng định nghĩa tính \(f'\left(x\right),x\in\left(3;+\infty\right)\)

b) Viết PT tiếp tuyến tại x₀=5

c) Giải bất phương trình \(f'\left(x\right)>4,x\in\left(3;+\infty\right)\)

Answer 1

\(\Delta y=4\sqrt{2\left(x+\Delta x\right)-6}-4\sqrt{2x-6}=\frac{8\Delta x}{\sqrt{2x+2\Delta x-6}+\sqrt{2x-6}}\)

\(f'\left(x\right)=\lim\limits_{\Delta\rightarrow0}\frac{\Delta y}{\Delta x}=\lim\limits_{\Delta x\rightarrow0}\frac{8\Delta x}{\Delta x\left(\sqrt{2x+2\Delta x-6}+\sqrt{2x-6}\right)}\)

\(=\lim\limits_{\Delta x\rightarrow0}\frac{8}{\sqrt{2x+2\Delta x-6}+\sqrt{2x-6}}=\frac{8}{2\sqrt{2x-6}}=\frac{4}{\sqrt{2x-6}}\)

b/ \(f'\left(5\right)=\frac{4}{\sqrt{2.5-6}}=2\) ; \(f\left(5\right)=4\sqrt{2.5-6}=8\)

Pt tiếp tuyến: \(y=2\left(x-5\right)+8=2x-2\)

c/ \(f'\left(x\right)>4\Leftrightarrow\frac{4}{\sqrt{2x-6}}>4\Leftrightarrow\frac{1}{\sqrt{2x-6}}>1\)

\(\Leftrightarrow\sqrt{2x-6}< 1\Leftrightarrow2x-6< 1\Rightarrow x< \frac{7}{2}\)

\(\Rightarrow3< x< \frac{7}{2}\)