Using the limit definition, how do you find the derivative of $f (x) = x^{2} - 5 x + 3$ $f(x)= x^2 -5x + 3$ ?

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Answer 1 · 2018-06-11T01:10:19

$f'(x) = 2x - 5$

Given: $f(x) = x^2 - 5x + 3$ use the limit definition of the derivative.

Limit definition of the derivative:

$f'(x) = lim h-> 0 " "(f(x+h) - f(x))/(h$

Use substitution to find $f(x + h):$

$f(x + h) = (x + h)^2 - 5(x + h) + 3$

$= x^2 + 2xh + h^2 -5x -5h + 3$

$f'(x) = lim h-> 0 " "(x^2 + 2xh + h^2 -5x -5h + 3 - (x^2 - 5x + 3))/h$

Distribute the negative:

$f'(x) = lim h-> 0 " "(cancel(x^2) + 2xh + h^2 cancel(-5x) -5h + cancel(3) cancel(- x^2) + cancel(5x) cancel(- 3))/h$

$f'(x) = lim h-> 0 " "(2xh + h^2 - 5h)/h$

Factor $h$ from the numerator:

$f'(x) = lim h-> 0 " "(cancel(h)(2x + h - 5))/cancel(h)$

$f'(x) = lim h-> 0 " "2x + h - 5$

Take the limit (let $h -> 0): f'(x) = 2x - 5$

Using the limit definition, how do you find the derivative of f(x)= x^2 -5x + 3f(x)=x2−5x+3f(x)= x^2 -5x + 3?