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.