Use the Intermediate Value Theorem to determine if h(x)= 1-3tanx has a zero in the interval [0,pi/4]. You must show how you arrived at your answer using the IVT.

Question

reemii · Answer

the thing is to find an $x_1$ and an $x_2$ such that $h(x_1)<0$ and $h(x_2)>0$. Then the theorem tells you that .....

experimentx · Answer

$$ 
f(x) = 1 - 3 \tan(x)  \
f(0) = ? \
f(\pi/4) = ??  $$

anonymous · Answer

With the two values f(0) = 1 and f(pi/4)= -2, what's the next step? (-2+1)/2=-.5?

reemii · Answer

you use immediately the theorem

reemii · Answer

thanks to the theorem ew know that there's an $x$ in the interval such that $f(x)=-1$. but that works for the value 0 too.. (any value between 1 and -2)

anonymous · Answer

ohh ok, thanks.