Sam9530130

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These are answers submitted by Sam9530130

Yes joe, I meant piecewise function, my bad!

 

> g := x-> piecewise(0<=x and x<=c, m1*x+b1, c<x and x<=1, m2*x+b2);

f := x-> x^3;

I can not use commands we have not seen such as convert, collect.

Found my error!

Hope that helps :

Integral is from 0 to 1 ( [0,1] )

C is a constant in the interval [0,1]

f(x) = x^3  and 

g(x) = 

m1x + b1   if  0 ≤ x ≤ c

m2x + b2   if  c < x ≤ 1

Question : For what value of C is the area between the curves minimal?

I have 2 functions, f(x) and g(x) both continous on [a,b] ( [a,b] being a known interval). For what value of c ε [a,b] is the area between f(x) and g(x) minimal?

-Both functions are known, I can provide them if needed.

I hope it is easier to understand!

Good, 4 equations will do then, thanks alot for the help, I really appreciate it guys!
Well I did what you guys told me, it worked with 4 equations. I wanna know how to do it with only 2 equations. Here is it: f(x) = -x^4+11*x^3-38*x^2+52*x-14
Thanks !
I dont need to know x1 and x2, only a and b. So I only need 2 equations. I tried something on maple and it didn't work. Which 2 equations can I use?
> solve(8=tan(m)*6-((9.8)/2*20^2*cos(m)^2)*6^2,m); 1.592353478 - 0.03882662969 I, -1.549239176 - 0.03882662969 I, 1.592353478 + 0.03882662969 I, -1.549239176 + 0.03882662969 I, -1.613910629, 1.527682025 I guess I need to chose 1.527682025 because it is the only value that respects the equation? edit : I'll have to convert the answer to get an angle but does it look good?
Alright I'll do that thanks!
Thanks so much guys!
The slope of the line y="a"*x+b it's a?
I'd say you have to find the slope at x1 with the derivate of f(x1) so it'd be f'(x1)=a*x+b
A line is tangent to a curve at a point, if both line and curve pass through the point with the same direction?
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