Answer
The interval of $x$ is $[3.3833,3.3870]$
Work Step by Step
The exercise challenges us to find the interval of $x$ on which the inequality $|A-9|\le0.01$ holds.
To find out, we just have to solve the inequality:
$$|A-9|\le0.01$$ $$\Big|\pi\Big(\frac{x}{2}\Big)^2-9\Big|\le0.01$$ $$-0.01\le\pi\Big(\frac{x}{2}\Big)^2-9\le0.01$$ $$8.99\le\pi\Big(\frac{x}{2}\Big)^2\le9.01$$ $$\frac{8.99}{\pi}\le\Big(\frac{x}{2}\Big)^2\le\frac{9.01}{\pi}$$
Since $x$ here is the diameter of the cylinder, $x\gt0$, and $x/2\gt0$ as well; therefore $\sqrt{(x/2)^2}=|x/2|=x/2$
Take the square root of all sides: $$\sqrt{\frac{8.99}{\pi}}\le\frac{x}{2}\le\sqrt{\frac{9.01}{\pi}}$$ $$2\sqrt{\frac{8.99}{\pi}}\le x\le2\sqrt{\frac{9.01}{\pi}}$$ $$3.3833\le x\le3.3870$$
Hence, for the inequality $|A-9|\le0.01$ to hold, $x\in[3.3833,3.3870]$
