# Is this right?

is this right??

1. 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)

2. Homework Equations
f(a)= f(x) - f(a)/x-f(a)

3. 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?

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HallsofIvy
Homework Helper
1. 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)

2. Homework Equations
f(a)= f(x) - f(a)/x-f(a)
Do you mean "f'(a)= (f(x)-f(a))/(x- a)" here?

3. 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.

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