Question

Show that e^xsinx = 0 has a solution in the interval [pi/2, 3pi/2]

I just have to show that it excists, but how do I do that?

Answer 1

\(e^x\sin x=0\) means either factor equals 0 .. think of when you solve \((x-a)(x-b)=0\)
\(e^x=0\) or \(\sin x =0 \), but \(e^x\) is never 0, so can you find the solution for \(\sin x = 0\) in that interval ?

Answer 2

i dont get it...

Answer 3

Let's say you have a quadratic that you are solving for: \(x^2-4=0\)
You can solve for x by factoring: (using a difference of squares):
\((x-2)(x+2)=0\)

Now that answer is 0 if: \((x-2)=0 \) or \((x+2)=0\)

Now apply this same idea to your equation:
\(e^x\sin x=0\\
(e^x)(\sin x)=0\)

This is equal to 0 if
\((e^x)=0 \) or \((\sin x) =0\)

But \(e^x\) i the exponential function, and there is NO \(x\) that will render \(e^x=0\) true. (If you that the exponential function, the function never crosses the x-axis!, it has no root a.k.a. no zero)
So the only way to make your equation 0 is if \(\sin x =0\)
and do you know any value of \(x\) that makes \(\sin x = 0 \) ?

Answer 4

If you look at the exponential function graphically*