Solving Mapping Problem: Let D = {x ∈ R, -3 ≤ x ≤ 5, x ≠ 0}

[gutnedawg]

Let D = {x $$\in$$ R : -3$$\leq$$ x $$\leq$$ 5 and x $$\neq$$0}
and define g(x) = [cos(x) - 1]/x + sqrt(x+3)(5-x)

Find G: R ->R such that G is continuous everywhere and g(x)=G(x) when x$$\in$$D

I'm not really sure how to start this and I've been looking at it for quite a while now. I might just need a push in the correct direction.

 

[Office_Shredder]

You need to do two things:

sqrt( (x+3)(5-x) ) is not defined when x is larger than 5 or smaller than -3. How can you modify it so that you have a function that is defined outside of that interval?

(cos(x)-1)/x is not defined at 0. I think this is the trickier part of the question, you need to deal with the removable discontinuity at zero.

 

[gutnedawg]

You need to do two things:

sqrt( (x+3)(5-x) ) is not defined when x is larger than 5 or smaller than -3. How can you modify it so that you have a function that is defined outside of that interval?

(cos(x)-1)/x is not defined at 0. I think this is the trickier part of the question, you need to deal with the removable discontinuity at zero.

I know I have to do this but my math is rather rusty and I'm just not seeing it