Andiio's Questions Thread :)

[Andiio]

Noob question:

f(x) = 1/(x+a)^2 and g(x) = √x, determine the values of a such that f[g(x)] exists.

SO dom f = R\{-a}, and dom f ≥ [0,+∞)
But then why is a > 0?

Thanks!

[nacho]

because x = a would then also be 0

[xZero]

the range of g(x) is [0,∞), for f[g(x)] to exist, the range of g(x) must be a subset of the domain of f(x). The range of g(x) tells us that f(x) must exist from 0 to ∞, hence if a is less than 0, there will be an asymptote between 0 and ∞. So from this we have to make a>0 so the asymptotes not within 0 and ∞. Hope it makes sense

[brightsky]

-a must not be in the interval [0, infty) or else the fog dies. So -a < 0, which gives a > 0.

[luffy]

Noob question:

f(x) = 1/(x+a)^2 and g(x) = √x, determine the values of a such that f[g(x)] exists.

SO dom f = R\{-a}, and dom f ≥ [0,+∞)
But then why is a > 0?

Thanks!

This question has already been explained, but I figured I would mention it anyway:

exists if .
Firstly, find the range of , which is
The domain of in general terms, is .

Now, you need the domain of , which is , to be greater than the range of , which is . If a < 0 (lets use -2 as an example), you will get , then the range of will contain a single point at x = 2, which is not in the domain of .
->Therefore, f[g(x)] will not exist.

Now, if a > 0, (lets use 2), then domain of , which will contain all values present in the range of , which is .  This would make .

Thus, the values of 'a' such that exists is  

[nacho]

You mean:

This question has already been explained, but I figured I would mention it anyway:

exists if .

[luffy]

You mean:

This question has already been explained, but I figured I would mention it anyway:

exists if .

typo - fixed

[Andiio]

What would be the quickest, and less tedious method of solving this question?

Find the coefficient of y^4 in the expansion of (y+3)^3(2-y)^5.

Thanks!

[brightsky]

We want to find the product of the coefficients of the y^3 in (y+3)^3 and y of (2-y)^5 plus the product of the coefficients of y^2 in (y+3)^3 and y^2 of (2-y)^5 plus the product of the coefficients of y in (y+3)^3 and y^3 of (2-y)^5 plus, finally, the product of the coefficient of y^0 in (y+3)^3 and y^4 in (2-y)^5.

Hence the coefficient is given by:
1*(-5*2^4) + (3*3)(10*2^3) + (3*3^2)*(-10*2^2) + (3^3)*(5*2) = -170