How do you use the limit definition to find the derivative of #f(x)=x^2-15x+7#?

1 Answer
Jan 2, 2017

Expand, reduce and evaluate the limit.

Explanation:

I'll use #f'(x) = lim_(hrarr0)(f(x+h)-f(x))/h#

Long version explanation

For #f(x) = x^2-15x+7#,

notice that #f(x+h) = (x+h)^2-15(x+h)+7#.

#f'(x) = lim_(hrarr0)(f(x+h)-f(x))/h#

# = lim_(hrarr0)(((x+h)^2-15(x+h)+7)-(x^2-15x+7))/h#

Notice that, if we try to evaluate by substitutiton, we get the indeterminate form #0/0#.

Expand the numerator:
(note #(x+h)^2 = x^2+2xh+h^2# #" "# use FOIL if you need to)

# = lim_(hrarr0)((x^2+2xh+h^2-15x-15h+7)-(x^2-15x+7))/h#

# = lim_(hrarr0)((x^2+2xh+h^2-15x-15h+7-x^2+15x-7))/h#

Now, some of the terms in the numerator add to #0#

# = lim_(hrarr0)((color(red)(x^2)+2xh+h^2 color(green)(-15x) -15hcolor(blue)(+7) color(red)(-x^2) color(green)(+15x)color(blue)(-7)))/h#

# = lim_(hrarr0)(2xh+h^2-15h)/h#

We still get #0/0#, but we can factor and reduce

# = lim_(hrarr0)(cancel(h)(2x+h-15))/cancel(h)_1#

# = lim_(hrarr0)(2x+h-15)#

And now we can evaluate the limit

# = 2x+(0)-15 = 2x-15#

So, #f'(x) = 2x-15#

Short version

#f'(x) = lim_(hrarr0)(f(x+h)-f(x))/h#

# = lim_(hrarr0)(((x+h)^2-15(x+h)+7)-(x^2-15x+7))/h#

# = lim_(hrarr0)((x^2+2xh+h^2-15x-15h+7-x^2+15x-7))/h#

# = lim_(hrarr0)(2xh+h^2-15h)/h#

# = lim_(hrarr0)(2x+h-15)#

# = 2x-15#