Question

Find the derivative using the limit definition f(x)= 2x^2 -4x + 3

Answer 1

Answer:

Given that to ind the derivative using the limit definition

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

we have that,

$f^{\prime}(x)=\lim _{h\to0}\frac{f(x+h)-f(x)}{h}$

we get,

$f\mleft(x+h\mright)=2(x+h)^2-4(x+h)+3$

Solving this we get,

$f(x+h)=2(x^2+2xh+h^2)-4x-4h+3$$f(x+h)=2x^2-4x+3+4xh+2h^2-4h$

Substitute the values in f'(x) is,

$f^{\prime}(x)=\lim _{h\to0}\frac{2x^2-4x+3+4xh+2h^2-4h-(2x^2-4x+3)}{h}$$f^{\prime}(x)=\lim _{h\to0}\frac{4xh-4h+2h^2}{h}$$f^{\prime}(x)=\lim _{h\to0}(4x-4+2h^{})$$f^{\prime}(x)=4x-4$

the derivative using the limit definition of f(x) is 4x-4

Answer is: 4x-4.