Maxima minina on an interval (calculus+trig)

  • Thread starter Thread starter quicksilver123
  • Start date Start date
  • Tags Tags
    Interval Maxima
quicksilver123
Messages
173
Reaction score
0
IMG_2066.jpg
I need to find the max/min of a function on an interval.
The function is f(x)=x+cos(x) and the interval is <-PI,2pi>
There is an attached solution but I do not understand how to arrive at the given solution (see screenshot). I would personally just take the derivative as
F'(x)=1-sin(x)
However the solution says this is only half the answer and I do not understand the reasoning (I am trying to do this algebraicly, without thinking of it graphically).
 
Physics news on Phys.org
quicksilver123 said:
the solution says this is only half the answer
I think you may be misinterpreting what it says. It is strangely worded. Better would be "f'(x)=1-sin(x), which is ≥0 since sin(x)≤1 for any real x".
 
  • Like
Likes quicksilver123
Thanks
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top