Answer to Question #244078 in Calculus for Princewill Unuigbe

Question

For the function f(x) = 2x − 3 x (a) Use the limit definition of the derivative to find f 0 (x). (b) Find the equation of the line tangent to f(x) at x = 3.

Answer 1

Given that $f(x)=2x-3$

(a) For $y=f(x)$ , the derivative of y (with respect to x ) is

 

$\begin{aligned} &y^{\prime}=\frac{d y}{d x}=\lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x} \\ &f^{\prime}(x)=\lim _{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x} \\ &f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ &f^{\prime}(x)=\lim _{u \rightarrow x} \frac{f(u)-f(x)}{u-x} \end{aligned}$

$\therefore f'(0)= \begin{aligned}\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ \end{aligned}$

$\Rightarrow f'(0)=\lim\limits_{h \rightarrow 0} \frac{2(x+h)-3-2x+3}{h} \\ \Rightarrow f'(0)=\lim\limits_{h \rightarrow 0} \frac{2h}{h}\\ \Rightarrow f'(0)=2$

(b) Given that $f(x)=2x-3$

$\therefore f'(x)=2\\ So, f'(3)=2$

At $x=3,$ we get: $y=6-3=3$

Equation of tangent:

$y-3=2(x-3)\\ \Rightarrow y=2x+3$