College exam calculus I

[poissonrouge]

Hi i don't understand how to answer this question :
The equation [MATH] x(e^{x}-e^{-x})-e^{x}=0 [/MATH]A. doesn't have any of its solutions in the interval [0,+[MATH] \infty [/MATH] ]
B. has exactly one solution in [MATH] \Bbb R [/MATH]C. doesn't have any of its solutions in the interval ]-[MATH]\infty[/MATH], 0[
D. has at least two solutions in [MATH] \Bbb R [/MATH]
I don't understand how to solve this (the answer is D.). Does anyone know? Thanks a lot

 

[tkhunny]

Yes. What have you tried?

Can you just graph it?
Can you find the minimum or maximum point? <== This, at least, looks like calculus.
Have you tried rephrasing the equation?
Maybe try a few values? Exponentials tend to take off pretty quickly, so stick pretty close to x = 0. What are the values of the left hand side for x = 0 and x = 1? Does that tell us anything? Try the negative side? You may have to go all the way to x = -2 to see this side.

 

[Steven G]

I would multiply both sides by e^x getting e^2x(x-1) = x. I would look at y = x and y = e^2x(x-1) and see if y=e^2x(x-1) is on both sides of x or always below or above x to answer this question.
This is not an easy problem. Are you allowed to use a graphing calculator?

 

[poissonrouge]

I would multiply both sides by e^x getting e^2x(x-1) = x. I would look at y = x and y = e^2x(x-1) and see if y=e^2x(x-1) is on both sides of x or always below or above x to answer this question.
This is not an easy problem. Are you allowed to use a graphing calculator?

I would multiply both sides by e^x getting e^2x(x-1) = x. I would look at y = x and y = e^2x(x-1) and see if y=e^2x(x-1) is on both sides of x or always below or above x to answer this question.
This is not an easy problem. Are you allowed to use a graphing calculator?

Thats what i did but i dont know what to do next. Im not allowed anything during the exam I guess there's an easy to tell or a theorem that can apply in this situation but i dont know..

 

[Dr.Peterson]

Hi i don't understand how to answer this question :
The equation [MATH] x(e^{x}-e^{-x})-e^{x}=0 [/MATH]A. doesn't have any of its solutions in the interval [0,+[MATH] \infty [/MATH] ]
B. has exactly one solution in [MATH] \Bbb R [/MATH]C. doesn't have any of its solutions in the interval ]-[MATH]\infty[/MATH], 0[
D. has at least two solutions in [MATH] \Bbb R [/MATH]
I don't understand how to solve this (the answer is D.). Does anyone know? Thanks a lot

I'm inclined to use the intermediate value theorem, perhaps starting with x=0 and limits as x approaches + or - infinity. These are suggested by options A and C, in particular.

 

[Steven G]

I agree with Dr Peterson about using the intermediate value theorem. I hinted that by asking if e^2x(x-1) is on both sides of y=x. I guess it would have been better if I asked how many times does e^2x(x-1) cross y=x.