Can the intermediate-value theorem be used to show there is a solution the equation f (x) = 0 on the interval [0, π⁄2 ]? Give an explanation why.
F(x) = 2sin(x) - 3cos(x) -x^2

Question

anonymous · Answer

a) No. because f (0) > 0 and f ( π⁄2 ) > 0. b) No, because f (0) < 0 and f ( π⁄2 ) < 0. c) Yes, because f (0) > 0 and f ( π⁄2 ) < 0. d) Yes. because f (0) < 0 and f ( π⁄2 ) > 0.

anonymous · Answer

replace x by 0 and also by 
$$\frac{\pi}{2}$$

anonymous · Answer

if one is positive and the other answer is negative, then the function must be zero somewhere in between.

anonymous · Answer

$$f(0)=2\sin(0)-3\cos(0)-0^2=0-3-0=-3$$

anonymous · Answer

just taking this all in.

anonymous · Answer

$$f(\frac{\pi}{2})=2-0-(\frac{\pi}{2})^2$$\]

anonymous · Answer

the  function is continuous yes?  so it does not skip values

anonymous · Answer

if it is say 3 at some point and then -2 at some other point it must take on all values in the interval from -2 to 3.  unfortunately in this case both answers are negative

anonymous · Answer

what is the 2nd value after plugging in pi/2?

anonymous · Answer

first plug in 0.  i got -3 did you?

anonymous · Answer

then plug in $$\frac{\pi}{2}$$

anonymous · Answer

yes

anonymous · Answer

i got 
$$2-(\frac{\pi}{2})^2$$

anonymous · Answer

i got both, since one is pos and the other neg it must cross the x axis

anonymous · Answer

actually no

anonymous · Answer

-3 is negative

anonymous · Answer

I got 2sin(pi/2/0 -3 cos(pi/2) -(pi/x)^2

anonymous · Answer

and so is $$2-(\frac{\pi}{2})^2$$

anonymous · Answer

$$F(x) = 2\sin(x) - 3\cos(x) -x^2$$
$$F(\frac{\pi}{2}) = 2\sin(\frac{\pi}{2}) - 3\cos(\frac{\pi}{2}) -(\frac{\pi}{2})^2$$

anonymous · Answer

$$=2-(\frac{\pi}{2})^2$$

anonymous · Answer

whoops missed the 2, for pi/2 for sin

anonymous · Answer

got it now!

anonymous · Answer

but this is negative also.  i just checked

anonymous · Answer

so both values are negative, so we don't know whether it crossed the axis or not

anonymous · Answer

yes it is

anonymous · Answer

got it

anonymous · Answer

answer b is correct.

anonymous · Answer

therefore IVT cannot explain

anonymous · Answer

zactly

anonymous · Answer

got another one, going to do it meow

anonymous · Answer

thank you, that was very clear!

anonymous · Answer

i'm ready.  let me get my beer

anonymous · Answer

haha!!!!