Question

f(theta)=sin theta cos^2 theta - (cot theta)/ theta + 1 Domain: [0, 2pi] Find: 1) what are the inflection point 2) relationship of stationary point and critical point 3) what are the critical points

Analyzing functions
ANSWERED
asked 2021-02-11
\(\displaystyle{f{{\left(\theta\right)}}}={\sin{\theta}}{{\cos}^{{2}}\theta}-\frac{{{\cot{\theta}}}}{\theta}+{1}\)
Domain: \(\displaystyle{\left[{0},{2}\pi\right]}\)
Find:
1) what are the inflection point
2) relationship of stationary point and critical point
3) what are the critical points

Answers (1)

2021-02-12
1) An inflection point is where the curve of the graph goes from concave down to up or vice versa. Point of inflection occur at f"=0
2) Relationship of stationary point and critical point.
We say \(\displaystyle{x}_{{0}}\) is a stationary point of a function if f(x) and f'(x) exist and is equal to \(\displaystyle{f}'{\left({x}_{{0}}\right)}={0}.\)
And, \(\displaystyle{x}_{{0}}\) is a critical point of a function of f(x) if \(\displaystyle{f{{\left({x}_{{0}}\right)}}}\) exists and either \(\displaystyle{f}'{\left({x}_{{0}}\right)}\) does not exist (i.e. function is not differentiable at or \(\displaystyle{f}'{\left({x}_{{0}}\right)}={0}.\)
All stationary points are critical points but not all critical points are stationary points.
3) Point \(\displaystyle{x}_{{0}}\) is a critical point of a function of f(x) if \(\displaystyle{f{{\left({x}_{{0}}\right)}}}\) exists and either \(\displaystyle{f}'{\left({x}_{{0}}\right)}\) does not exist (i.e. function is not differentiable at \(\displaystyle{x}_{{0}}\) or \(\displaystyle{f}'{\left({x}_{{0}}\right)}={0}.\)
The first derivative test is a method of analyzing functions using their first derivatives in order to find their extremum point.
We take the derivative of the function,
\(\displaystyle{f}'{\left(\theta\right)}=\frac{{d}}{{{d}\theta}}\cdot{\left({\sin{\theta}}{{\cos}^{{2}}\theta}-\frac{{{\cos{\theta}}}}{\theta}+{1}\right)}\)
\(\displaystyle=\frac{{d}}{{{d}\theta}}\cdot{\left({\sin{\theta}}{{\cos}^{{2}}\theta}\right)}-\frac{{d}}{{{d}\theta}}{\left(\frac{{{\cos{\theta}}}}{\theta}\right)}+\frac{{d}}{{{d}\theta}}{\left({1}\right)}\)
\(\displaystyle={{\cos}^{{3}}\theta}-\frac{{-\theta{\cos{{e}}}{c}^{{2}}\theta-{\cot{\theta}}}}{\theta^{{2}}}-{\sin{{2}}}\theta{\sin{\theta}}\)
Now, substitute \(\displaystyle\theta={4}\)
\(\displaystyle{f}'{\left({0}\right)}=-{0.27927}-{\left(-{0.49047}\right)}-{\left(-{0.74875}\right)}={0.959951}\stackrel{\sim}{=}{0.96}\)
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