Question

How do I solve for the points of inflection involving trig functions?

Answer 1

In this question, we have to find the point of inflection. As we know, an inflection point is a point of a curve at which there is a change in the direction of curvature. Also, points of inflection on a graph is where the concavity of the graph changes. Thus, in this question we will first find the differentiation of the function. After that, we will again differentiate the first differentiate function and then put it equal to 0, to get the value of the variable. After that, we will apply the quadrant rule, because the given function is of trigonometric function. Then, we will make the necessary calculations, to get the result.

Complete step by step solution:
According to the question, we have to find the points of inflection. As we know, a point of inflection shows the concavity of the curve, that is when we put the value of second differentiation equal to 0, we get the point of inflections.
Thus, we let a trigonometric function for the given problem. Let us suppose a function $f(x)=\sin x+\cos x$ with the interval $[0,2\pi ]$ --------- (1)
So, first we will find the differentiation with respect to x of the equation (1), that is differentiation of sinx is cosx and cosx is –sinx, therefore we get
$\Rightarrow {f}'(x)=\cos x-\sin x$
Now, we will again differentiate the above equation with respect to x, we get
$\Rightarrow {f}''(x)=-\sin x-\cos x$
Now, we will put the value of second derivative equal to 0, therefore we get
$\Rightarrow -\sin x-\cos x=0$
Now, we will add cosx on both sides in the above equation, we get
$\Rightarrow -\sin x-\cos x+\cos x=0+\cos x$
As we know, the same terms with opposite signs cancel out each other, thus we get
$\Rightarrow -\sin x=\cos x$
Now, we will divide cosx on both sides in the above equation, we get
$\Rightarrow -\dfrac{\sin x}{\cos x}=\dfrac{\cos x}{\cos x}$
On further solving, we get
$\Rightarrow -\tan x=1$
Now, we will multiply (-1) on both sides in the above equation, we get
$\Rightarrow -\tan x.\left( -1 \right)=1\left( -1 \right)$
Therefore, we get
$\Rightarrow \tan x=-1$
Now, we know that tan function is minus 1 when the angle is equal to $\dfrac{3\pi }{4}+n\pi$ , that is
$\Rightarrow x=\dfrac{3\pi }{4}+n\pi$
$\Rightarrow x=\dfrac{3\pi }{4},\dfrac{3\pi }{4}+\pi$
On further simplification, we get
$\Rightarrow x=\dfrac{3\pi }{4},\dfrac{7\pi }{4}$
Therefore, we get the intervals $\left( 0,\dfrac{3\pi }{4} \right),\left( \dfrac{3\pi }{4},\dfrac{7\pi }{4} \right),\left( \dfrac{7\pi }{4},2\pi \right)$ ------- (2)
Now, we will let one point of both the intervals and put it in the value of the function. Thus, if we get the value
Let us suppose a point from the interval $\left( 0,\dfrac{3\pi }{4} \right)$ be $\dfrac{\pi }{2}$ . Thus we will substitute this value in the given function that is in equation (1), we get
$\Rightarrow f\left( \dfrac{\pi }{2} \right)=\sin \left( \dfrac{\pi }{2} \right)+\cos \left( \dfrac{\pi }{2} \right)$
On further solving, we get
$\Rightarrow f\left( \dfrac{\pi }{2} \right)=1+0$
$\Rightarrow f\left( \dfrac{\pi }{2} \right)=1$
Thus, the value of function at $\dfrac{\pi }{2}$ is an increasing function.
Now, let a point from the interval $\left( \dfrac{3\pi }{4},\dfrac{7\pi }{4} \right)$ be $\dfrac{3\pi }{2}$ . Thus we will substitute this value in the given function that is in equation (1), we get
$\Rightarrow f\left( \dfrac{3\pi }{2} \right)=\sin \left( \dfrac{3\pi }{2} \right)+\cos \left( \dfrac{3\pi }{2} \right)$
On further solving, we get
$\Rightarrow f\left( \dfrac{3\pi }{2} \right)=-1+0$
$\Rightarrow f\left( \dfrac{3\pi }{2} \right)=-1$
Thus, the value of function at $\dfrac{3\pi }{2}$ is a decreasing function.
So, we will also let a point from the interval $\left( \dfrac{7\pi }{4},2\pi \right)$ be $\dfrac{11\pi }{6}$ . Thus we will substitute this value in the given function that is in equation (1), we get
$\Rightarrow f\left( \dfrac{11\pi }{6} \right)=\sin \left( \dfrac{11\pi }{6} \right)+\cos \left( \dfrac{11\pi }{6} \right)$
On further solving, we get
$\Rightarrow f\left( \dfrac{11\pi }{6} \right)=-0.5+0.8$
$\Rightarrow f\left( \dfrac{11\pi }{6} \right)=0.3$
Thus, the value of function at $\dfrac{11\pi }{6}$ is an increasing function.
Therefore, the points of inflection for the trigonometric function $f(x)=\sin x+\cos x$ with the interval $[0,2\pi ]$ is $\dfrac{3\pi }{4},\dfrac{7\pi }{4}$ .

Note:
While solving this problem, do the step-by-step calculations to avoid confusion and mathematical error. Always remember to find the inflection point, we get the value of x from the second derivative and put it into the function, to get the concavity. Also, in trigonometric function, always remember the interval given in the problem, that is the points of inflection always lie in the given interval.