Question

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?

Answer 1

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`