Bump, still need help
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!
Last edited by goodieboy; 12-10-2012 at 08:20 PM.
Bump, still need help
@Reese, help me? Thanks! Will + rep.
Still need this done
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
Last edited by Graff; 12-09-2012 at 10:52 PM.
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)
Last edited by Reese; 12-09-2012 at 11:04 PM.
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.
Last edited by Graff; 12-10-2012 at 09:56 PM.