# Show equation has exactly one root

1. Jun 30, 2006

### darius

The question is show that the equation f(x)=3x-2+cos(Piex/2)=0 has exactly one real root.
Using the intermediate value theorem f is continuous and f(o)=-1 and f(1)=1., the equation has one root. Supposing it has two roots a and b , and
a<b. Then f(a)=0=f(b). Using Rolle's theorem there is a number d in (a,b) such that f'(x)=3-(pie/2)(sin(piex/2))=0 has a root in (a,b). But this is impossible since after factoring sin(piex/2)=6/pie is always greater than zero for all real x. I am having problems with factoring. Why is 3-(pie/2)(sin(piex/2))always >o for all x?

2. Jun 30, 2006

### d_leet

First it's pi not pie. Now for your question as to why it is always greater than zero this is because you reduced this to sin(pi*x/2) = 6/pi, but the range of the sine function is only from -1 to 1, and you should notice that 6/pi is greater than 1 hence 3 - (pi/2)(sin(pi*x/2)) is always greater than 0 for real x.

3. Jun 30, 2006

### darius

thank you so much. Sometimes, we have a hard time seeing the easiest step after going though the hard steps!! I am just beginning calculus and I am starting college in the fall. Thank you!! P. S. today was my first day at this forum!

4. Jun 30, 2006

### d_leet

Yes I know exactly what you mean, I'm glad i could help and welcome to the forums.