Question

How do you use the intermediate value theorem to verify that there is a zero in the interval [0,1] for `h(theta)=1+theta-3tantheta`?

Answer 1

See explanation.

Explanation:

`theta` in radian measure befits the question.

So, `theta in (0, 1)=(0, 57.3^o)`

`h(theta)` is continuous and differentiable in the broader #[0,

pi/2)#.

`h(0) = 1 > 0 and h(1)=1+1-3 tan 1=-2.473 < 0`

Therefore, there exists at least one zero in the interval (0, 1)