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

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Answer 1 · 2016-11-15T16:51:03

See explanation.

$θ$ $theta$ in radian measure befits the question.

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

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

pi/2)#.

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

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

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