Question

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

Answer 1

Please see the explanation.

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]"`