I have a problem with this exercise. This is the text. The function $$f(x) = x-\ln (1 + 2x ^ 2)$$ in the interval $[1,3]$ has:

  1. two points of maximum and a minimum point relative or absolute

  2. a minimum point and no point of maximum relative or absolute

  3. a maximum point and a minimum point relative or absolute

  4. a maximum point and two points of minimum relative or absolute

I'm uncertain between the first and the second, this is the doubt: the points $f (1)$ and $f (3)$ should be inserted as the maximum (then right the first response) or not (then the second right)? this is the function in the range $[1,3]$

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The function is continuous in the interval $[1,3]$, hence by Weierstrass theorem it gets absolute maximum and absolute minimum in the interval.

Deriving the function we get $\displaystyle f'(x)=1-\frac{4x}{1+2x^2}=\frac{2x^2-4x+1}{1+2x^2} = 0$.

Solutions are $\displaystyle x_{1,2}=1\pm{\frac{1}{\sqrt{2}}}$. Only $\displaystyle 1+\frac{1}{\sqrt{2}}$ is in the interval. Using second derivative test We find that this is a minimum point, hence the function has 3 critical points.

In order to find which is relative and which is absolute, find the corresponding values of each critical point.

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  • $\begingroup$ i don't get why the $f(1)$ point is considered as minimum point? $\endgroup$ – user12 Jan 2 '16 at 19:37
  • $\begingroup$ @Amarildo, your function is defined in closed interval and it is continuous, hence it has critical points at end points of interval, in this case 1 and 3. $\endgroup$ – Galc127 Jan 2 '16 at 20:39
  • $\begingroup$ ok but why $f(1)$ is considered as minimum and not a maximum? this is what i dont get $\endgroup$ – user12 Jan 3 '16 at 13:17
  • $\begingroup$ @Amarildo, you are correct, $x=1$ is indeed a maximum - the function is decreasing in the interval $\left(1,1+\frac{1}{\sqrt{2}}\right)$ and then it is increasing, hence $x=1$ is maximum point, $1+\frac{1}{\sqrt{2}}$ is minimum point and $x=3$ is maximum point. Which is relative and which is absolute? $\endgroup$ – Galc127 Jan 3 '16 at 14:21
  • $\begingroup$ the absolute is the $x=3$, right? $\endgroup$ – user12 Jan 3 '16 at 14:27

Note the difference between relative maximum and absolute maximum.

Relative maximum may occur at a point where $f'(x) = 0$. You use the derivative of a function $f(x)$ to find relative maximum.

In order to find an absolute maximum, you need find the maximum value of a function $f(x)$ in an given interval $[a, b]$.

In your problem, you have an interval $[1, 3]$. The absolute maximum or minimum of your function is the maximum and minimum value in that interval.

The relative maximum or minimum occurs at points where your derivative $f'(x) = 0$. Make sure the point is in the given interval.

Can you use these ideas to find out what the answer is?

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  • $\begingroup$ so the answer is the second? $\endgroup$ – user12 Jan 2 '16 at 19:26
  • $\begingroup$ @Amarildo look at the other answer for help. $\endgroup$ – Varun Iyer Jan 2 '16 at 19:27

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