got a few problems im struggling with, any help is greatly appreciated

#1 the function is defined by f(x)= radical(25-x^2) for -5<x<5

now: let g be the function defined by g(x)= {f(x) for -5<x<-3
and {x+7 for -3<x<5

is g continuous at x=-3? use definition of continuity to explain answer



im so confused.

for a function to be continuous, you must have:
1.) the left hand and right hand limits must exist and be the same value
2.) the function value must equal the limit value

If you evaluate the limit for both of your piece-wise functions, you see a limit exists and is the same value for both (limit=4)
However, your domain of g(x) does not include -3, so f(-3) does NOT equal the limit value.
So g(x) is NOT continuous.

thank you, i think that cleared some things up. what makes you so sure the domain of g(x) doesnt include -3?

you have the domain listed.
-5<x<-3
and
-3<x<5

If you combine both of your domains, you have:
(-5,-3) union (-3,5)

So clearly, -3 is not in your domain

sorry its actually -5<x<-3
and -3<x<5

does that change the answer?

yes.
Since g(-3) exists, it equals 4
So the limit at -3 equals g(-3). This means the function is continuous

got it

now area of square is A=s^2

if area is 4 square inches and its side length is increasing at a rate of .5 inch per second, at what rate is the area changing? with respect to time (t seconds)

probably one of the main concept of calculus is the derivative

guys, why do you use the 'radical' when you actually mean 'square root'?
in mathematics a radical has a very clear definition in those areas the term is used but square root is not one of them