## Calculus (3rd Edition)

There is a zero in the interval of length $0.025$ $(0.575,0.6)$
We are given the function: $f(x)=\tan^3\theta-8\tan^2\theta+17\tan\theta-8$ Compute the value of $f$ for $x=0.5$ and $x=0.6$: $f(0.5)=\tan^3 0.5-8\tan^2 0.5+17\tan 0.5-8\approx -0.93739$ $f(0.6)=\tan^3 0.6-8\tan^2 0.6+17\tan 0.5-8\approx 0.20619$ As $f(0.5)f(0.6)<0$, $f$ has a zero in the interval of length $0.1$ $(0.5,0.6)$. The midpoint of the interval $(0.5,0.6)$ is $0.55$. Compute $f$ in $x=0.55$: $f(0.55)=\tan^3 0.55-8\tan^2 0.55+17\tan 0.55-8\approx -0.35393$ As $f(0.5)f(0.55)<0$, there is a zero in the interval of length $0.05$ $(0.55,0.6)$. The midpoint of the interval $(0.55,0.6)$ is $0.575$. Compute $f$ in $x=0.575$: $f(0.575)=\tan^3 0.575-8\tan^2 0.575+17\tan 0.575-8\approx -0.07078$ As $f(0.575)f(0.6)<0$, there is a zero in the interval of length $0.025$ $(0.575,0.6)$.