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It is bounded between $\pi/2$ and $2\pi$.

Is the mean value theroem involved in solving this? Any help will be appreciated. Thanks in advance.

Bobson Dugnutt
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21rw
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  • It clearly has no global max as it is unbounded above. Also min at $x=0$. – Zain Patel May 03 '17 at 01:01
  • @Zain Sorry I forgot to add the range. – 21rw May 03 '17 at 01:04
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    @Umoko Global extrema can only occur at stationary points or endpoints of the domain. Since the derivative is $>0$ for $x>\pi/2$ then you'll find that the extrema occur at the endpoints. – Zain Patel May 03 '17 at 01:06

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For $x>0$ the function is monotonically increasing. You can check this by differentiation and noting that there are no zeros in that range and that a bigger value of $x$ (for instance $2\pi$) leads to a bigger function-value.

Thus the minimum is at $\pi/2$ and the maximum is at $2\pi$.

Bobson Dugnutt
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HINT: All minimums/maximums must be critical points. Critical points are where the derivative is equal to $0$. I don't think you need the mean value theorem for this.

Start by finding the derivative for the function $y = 2x^2 + \cos(2x)$.

Set $y' = 0$, and solve for x.

Then, you can plug those values (if there are multiple) back in to your first equation to find your maximum and minimum. Note that intuitively you can figure out that there is no maximum.

John Lou
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