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

1 Answer
Jan 3, 2017

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]"#