# How do you use the intermediate value theorem to verify that there is a zero in the interval [0,1] for g(t)=2cost-3t?

There is no zero for $g \left(t\right) = 2 \cos t - 3 t$ in the range $\left[0 , 1\right]$.
It is observed that both $\cos t$ and $3 t$ are continuous in the range $\left[0 , 1\right]$, and hence $g \left(t\right) = 2 \cos t - 3 t$ is also continuous over the range $\left[0 , 1\right]$.
Now g(0)=2cos0-3×0=2-3=-1 and g(1)=2cos1-3=2×0.5403-3=-1.9194.
As $g \left(t\right)$ is continuous but does not change sign between $\left[0 , 1\right]$, we do not have a zero in the range $\left[0 , 1\right]$. In fact as the derivative of $g \left(t\right)$, $g ' \left(t\right) = - 2 \sin t - 3$ is negative in the range $g \left(t\right)$ is continuously decreasing in the range and never reaches the value $0$.