Using the limit definition, how do you find the derivative of $f (x) = 4 + x - 2 x^{2}$ $f(x)=4+x-2x^2$ ?

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Answer 1 · 2017-01-03T03:43:19

Please see the explanation.

The limit definition of the derivative is:

$f'(x) = lim_(hto0)(f(x+h) - f(x))/h" [1]"$

We are given that $f(x) = 4 + x - 2x^2" [2]"$

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

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

$f(x+h) = 4 + x + h - 2x^2 - 4hx - 2h^2" [3]"$

Substitute the right sides of equations [3] and [2] into equation [1]:

$f'(x) = lim_(hto0)((4 + x + h - 2x^2 - 4hx - 2h^2) - (4 + x - 2x^2))/h" [4]"$

Combine like terms in the numerator:

$f'(x) = lim_(hto0)(h - 4hx - 2h^2)/h" [5]"$

There is a common factor of h in the numerator and denominator:

$f'(x) = lim_(hto0)h/h(1 - 4x - 2h)" [6]"$

$h/h = 1$ , therefore, it disappears:

$f'(x) = lim_(hto0)(1 - 4x - 2h)" [7]"$

Now it is ok to let $hto0$ :

$f'(x) = 1 - 4x" [8]"$

Using the limit definition, how do you find the derivative of f(x)=4+x-2x^2f(x)=4+x−2x2f(x)=4+x-2x^2?