# 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?

Nov 15, 2016

See explanation.

#### Explanation:

$\theta$ in radian measure befits the question.

So, $\theta \in \left(0 , 1\right) = \left(0 , {57.3}^{o}\right)$

$h \left(\theta\right)$ is continuous and differentiable in the broader [0,

pi/2).

$h \left(0\right) = 1 > 0 \mathmr{and} h \left(1\right) = 1 + 1 - 3 \tan 1 = - 2.473 < 0$

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