How do you use the limit definition to find the slope of the tangent line to the graph #f(x)= 2x²-3x-5# at x=2?

1 Answer
Oct 30, 2016

The gradient of the slope when #x=2# is #m=5 #

Explanation:

By definition, # f'(x)= lim_(h->0)(f(x+h)-f(x))/h#

so for # f(x) = 2x^2 -3x -5 # we have:
# f'(x) = lim_(h->0)( { (2(x+h)^2 -3(x+h) -5) - (2x^2 -3x -5)} )/h #
# f'(x) = lim_(h->0)( (2(x^2+2hx+h^2) -3x -3h-5 ) - 2x^2+3x+5)/h#
# f'(x) = lim_(h->0)( 2x^2+4hx+2h^2 -3x -3h-5 - 2x^2+3x+5 )/h#
# f'(x) = lim_(h->0)( 4hx+2h^2-3h )/h#
# f'(x) = lim_(h->0)( 4x+2h -3)#
# f'(x) = 4x-3#

so when # x=2 => f'(x)=4(2)-3 = 8-3=5 #

Hence, The gradient of the slope when #x=2# is #m=5 #

You could equally use the definition to find #f'(2)# directly rather than #f'(x)#
# f'(x)= lim_(h->0)(f(x+h)-f(x))/h #
# => f'(2)= lim_(h->0)(f(2+h)-f(2))/h#