# Thread: Math homework help (should be easy for you cos I'm noob)

1. ## Math homework help (should be easy for you cos I'm noob)

It's the basics of Calculus, I think, need to use some differentiation

Find the range of values of 'a' such that the gradient of the tangent to the curve y = (2x + a)/(4 - x^2) is always positive. ('a' is a constant here)

I could get up to the point of dy/dx = (2x^2 + 2ax + 8)/((4 - x^2)^2), but can't go on after that.

All help appreciated!

2. Bump, still need help

3. @Reese, help me? Thanks! Will + rep.

Still need this done

4. The derivative you found is the slope. The question is asking about what ranges the slope is positive.

Is it from -8 to positive infinity?

y = (2x + a)/(4 - x)

y prime= [(4 - x)(2) -(2x + a)(-1)] / (4 - x)^2

y prime = 8-2x+2x+a/(4 - x)^2

y prime = 8+a/(4 - x)^2 //denominator doesn't matter because it is squared. always positive except for a undefined at x=4

8+a>0

a>-8 <<---that's my conclusion

5. Heh, I started writing it out but Graff beat me to it . I can go and doublecheck, but everything looks good ^_^
I can try and explain everything if you'd like(gradient of tan=y2-21/x2-x1, there's lots of ways to look at what values are positive and what are negative, and that's usually just what makes sense to you)

6. Dammit I made a friggin mistake when typing the question. Will + full rep to both of you if you guys can help me solve it again. I've edited the question, help me again please?

@Graff @Reese, the answer should be -4 < a < 4

7. y = (2x + a)/(4 - x^2)

y prime= [(4 - x^2)(2) -(2x + a)(-2x)] / (4 - x^2)^2

y prime= 2(x^2 + xa + 4)/(4-x^2)^2

2(x^2 + xa + 4)/(4-x^2)^2>0

[2(x^2 + xa + 4)/(4-x^2)^2>0]*(4-x^2)^2
[2(x^2 + xa + 4)>0]/2

(x^2 + xa + 4)>0

use quad formula (to find critical points)

{-a(+/-)sqrt[a^2-(4*1*4)] }/2

(a^2 -16)^.5 --->> [(a+4)(a-4)]^.5

CPs @ a=-4 and 4

yprime = 2(x^2 + xa + 4)/(4-x^2)^2

plug in 3 values that are in between +/- infinity and the CP's to make sure that it's positive in that range only.

Struggled a bit, sorry. This is my final answer, I'll let you do the algebra.

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