+1904 Find the absolute maximum and absolute minimum values of f in the given interval.
f(t)=t+cot(t/2) [pi/4, 7pi/4]
Please show how you got your answer by finding the critical numbers by finding the derivative of the equation first and then setting the first derivative equal to zero. Then find the points where t=pu/4 and t=7pi/4.
+118687 Hi Gibson338 :)
Find the absolute maximum and absolute minimum values of f in the given interval.
f(t)=t+cot(t/2) [pi/4, 7pi/4]
pi/4 = 0.7853981633974483
7pi/4 = 5.4977871437821382
end points
f(pi/4)=pi/4+cot(pi/8) = pi/4+cot(pi/8) = 3.20
f(7pi/4)= 7pi/4+cot(7pi/8) = 7pi/4+cot(7pi/8) = 3.08
end points function values are approx 3.2 and 3.08
\(\text{Firstly cot(t/2) must be defined}\\ so\;\; sin(t/2) \ne0\\ so \;\;t/2\ne n\pi \qquad where\;\;n\in Z\\ so \;\;t\ne 2n\pi \qquad where\;\;n\in Z\)
\(f(t)=t+cot(t/2)\\ let\;\;y=f(t)\\ y=t+\frac{cos(t/2)}{sin(t/2)}\\ let\\ u=cos(t/2) \qquad \\ \frac{du}{dt}=0.5*-sin(t/2)\\ \frac{du}{dt}=-0.5sin(t/2)\\~\\ v=sin(t/2)\\ \frac{dv}{dt}=0.5*cos(t/2)\\~\\ \frac{dy}{dt}=1+\frac{vu'-uv'}{v^2}\\ \frac{dy}{dt}=1+\frac{sin(t/2)[-0.5sin(t/2)]-cos(t/2)0.5(cos(t/2)}{sin(t/2)^2}\\ \frac{dy}{dt}=1+\frac{-0.5sin^2(t/2)-0.5(cos^2(t/2)}{sin(t/2)^2}\\ \frac{dy}{dt}=1+\frac{-0.5}{sin(t/2)^2}\\\)
Find stat points that is when dy/dt = 0
\(0=1+\frac{-0.5}{sin(t/2)^2}\\ 1=\frac{0.5}{sin(t/2)^2}\\ sin(t/2)^2=0.5\\ sin(t/2)=\pm\sqrt{0.5}\\ \frac{t}{2}= n\pi \pm \frac{\pi}{4}\\ t= 2\pi n\pm \frac{\pi}{2}\\ \)
but
\(\frac{\pi}{4} \le t \le \frac{7\pi}{4}\\ t=\frac{\pi}{2}\;\;or \;\;\frac{3\pi}{2}\\ \text{So the stationary points are }\\ (\frac{\pi}{2},2.571),\quad (\frac{3\pi}{2},3.712) \)
\(end\; points\; ( \frac{\pi}{2}, 3.20) \;\;and \;\; ( \frac{7\pi}{4}, 3.08) \)
So the absolute minimum is approx 2.571 when t=pi/2
and the absolute maximum is approx 3.712 when t=3pi/2
Here is the graph:
