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# 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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