Question

How do you use the limit definition to find the slope of the tangent line to the graph `y=7x^3 - 2x^2 + 3x - 5` at x=-1?

Answer 1

Slope of tangent is `28`

Explanation:

The definition of the derivative of `y=f(x)` is

`f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h`

So with `f(x) = 7x^3-2x^2+3x-5` then;

`f(x+h) = 7(x+h)^3-2(x+h)^2+3(x+h)-5`

`" " = 7(x^3+3x^2h+3xh^2+h^3)-2(x^2+2xh+h^2)+3(x+h)-5`

`" " = 7x^3+21x^2h+21xh^2+7h^3 -2x^2-4xh-2h^2+3x+3h-5`

And so the limit numerator is:

`f(x+h)-f(x) = 7x^3+21x^2h+21xh^2+7h^3 -2x^2-4xh-2h^2+3x+3h-5-7x^3+2x^2-3x+5`

`" " = 21x^2h+21xh^2+7h^3 -4xh-2h^2+3h`

And so the derivative of `y=f(x)` is given by:

`f'(x) = lim_(h rarr 0) (21x^2h+21xh^2+7h^3 -4xh-2h^2+3h)/h`

`" " = lim_(h rarr 0) (21x^2+21xh+7h^2 -4x-2h+3)`

`" " = 21x^2+0+0 -4x-0+3`

`" " = 21x^2-4x+3`

And finally, the slope of the tangent at `x=-1` is given by `f'(-1)`, and:

`f'(-1) = 21+4+3 = 28`

Answer 2

`28.`

Explanation:

It is known that, the slope o the tgt. line to the graph,

`y=f(x)=7x^3-2x^2+3x-5,` at `x=-1,` is given by,

`[dy/dx]_(x=-1)=f'(-1).`

Now, using the Limit Defn. of `f'(-1),` we have,

`f'(-1)=lim_(x to -1) (f(x)-f(-1)}/{x-(-1)}.`

Here, `f(-1)=-7-2-3-5=-17.`

`:. f'(-1)=lim_(x to -1)[(7x^3-2x^2+3x-5)-(-17)}/(x+1),`

`=lim_(x to -1)(7x^3-2x^2+3x+12)/(x+1).`

In the poly. `7x^3-2x^2+3x+12,` since,

The sum of the co-effs. of odd-powers of `x,` i.e., `7+3=10,` is the

same as those of even-power, i.e., `-2+12=10,` we find that,

`(x+1)` is a factor of the poly.

Now, `7x^3-2x^2+3x+12,`

`=ul(7x^3+7x^2)-ul(9x^2-9x)+ul(12x+12),`

`=7x^2(x+1)-9x(x+1)+12(x+1),`

`=(x+1)(7x^2-9x+12),` so that,

`f'(-1)=lim_(x to -1) {(x+1)(7x^2-9x+12)}/(x+1),`

`=lim_(x to -1)(7x^2-9x+12),`

`=7-9(-1)+12=28,` is the reqd. slope of tgt.

Enjoy Maths.!