Find Slope of Graph at (-1,3) with Limit Process

[Precal_Chris]

is this right??

Homework Statement

use the limit process to find the slope of the graph of the function at the specified point.

f(x)= (sqroot of (x + 10)), at (-1, 3)

Homework Equations

f(a)= f(x) - f(a)/x-f(a)

The Attempt at a Solution

Lim as x-> -1 = sqrt of (x + 10) - sqrt of (-1 + 10)/ x- (-1)

Lim as x-> -1 = sqrt of (x + 10) - sqrt of (9) / x - (-1)

Lim as x-> -1 = sqrt of (x + 10) - 3 / x + 1

thats as far as i could go...
so is that the slope? or do i need to do nething further?

 

[HallsofIvy]

Homework Statement

use the limit process to find the slope of the graph of the function at the specified point.

f(x)= (sqroot of (x + 10)), at (-1, 3)

Homework Equations

f(a)= f(x) - f(a)/x-f(a)

Do you mean "f'(a)= (f(x)-f(a))/(x- a)" here?

The Attempt at a Solution

Lim as x-> -1 = sqrt of (x + 10) - sqrt of (-1 + 10)/ x- (-1)

Lim as x-> -1 = sqrt of (x + 10) - sqrt of (9) / x - (-1)

Lim as x-> -1 = sqrt of (x + 10) - 3 / x + 1

thats as far as i could go...
so is that the slope? or do i need to do nething further?

Yes, you need to do something further! You haven't taken the limit yet.

So far you have lim (x->-1) (sqrt(x+10)- 3)/(x+1) (Please use parentheses! It makes things much clearer!) Now, I would recommend "rationalizing the numerator": multiply both numerator and denominator by sqrt{x+ 10}+ 3. You should be able to cancel an "x+1" term in both numerator and denominator and then evaluate at x= -1.

 

[Precal_Chris]

Do you mean "f'(a)= (f(x)-f(a))/(x- a)" here?



Yes, you need to do something further! You haven't taken the limit yet.

So far you have lim (x->-1) (sqrt(x+10)- 3)/(x+1) (Please use parentheses! It makes things much clearer!) Now, I would recommend "rationalizing the numerator": multiply both numerator and denominator by sqrt{x+ 10}+ 3. You should be able to cancel an "x+1" term in both numerator and denominator and then evaluate at x= -1.

yes i did mean that...
and thnx i think i got it now.. but just to check ..is this right?

you have:
lim (x -> -1) = (sqrt(x+10)-3)/(x + 1) • (sqrt(x+10)+3)/(sqrt(x+10)+3)

then you have:

lim (x -> -1) = ((x+10)-9)/(x+1)(sqrt(x+10)+3)

(x+10)-9 simplifies to x+1 of course..then you can take the x+1's out of the numerator and denominator. then you get:

1/(sqrt(x+10)+3)
then when you put in the limit for x you get:

1/(sqrt(-1+10)+3)
1/(sqrt(9)+3
1/3+3
1/6

that is the answer?
[slope at (-1,3) is 1/6]

 

[Snazzy]

When I differentiate that function, I get the same answer; so yes, it seems right.