Let $f$ be the function with first derivative defined by $f'(x)=\sin(x^3)$ for $0\leq x\leq 2$. At what value of $x$ does $f$ attain its maximum value on the closed interval $0\leq x\leq 2$ ?

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Plot the graph in the graphing calculator to find intervals where the function is increasing.

Use your graphing calculator to obtain the intercepts. From the graph, $f'$ is positive from $0\lt x \lt 1.465$ and $1.845 \lt x \lt 2.112$.

We will therefore evaluate these two integrals to find the absolute maximum:

$$\int\limits_0^{1.465} f'(x)   dx \hskip{2em} \int\limits_0^{2} f'(x)   dx$$

Using the calculator:

$$\int\limits_0^{1.465} f'(x)   dx \approx 0.591$$ $$\int\limits_0^{2} f'(x)   dx \approx 0.452$$

Therefore, at $x=\boxed{1.465}$, $f$ reaches a maximum value.