Question

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

Answer 1

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)(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)(2x^2-4x+3+4xh+2h^2-4h-(2x^2-4x+3))/(h)`
`f^(\prime)(x)=\lim _(h\to0)(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.