6.2 Point of Inflection

  Monotonicity, Maxima and Minima

• The points in graph at which $f'(x) = 0{\text{ or }}f'(x) = {\text{does not exist}}$ are called the critical points. We use these critical points to find the points of inflection. The point must lie with in the domain of the function.
• In terms of concavity and convexity, we can say that the point $(c,f(c))$ on the graph $y=f(x)$ is the point of inflection if $f(x)$ is concave for the values less than $'c'$ and convex for values greater than $'c'$ or vice-versa.
• In other words, $f''(x)$ has opposite sign on either sides of $'c'$.
• If $f''(c)=0$ and $f'''(c) \ne 0$, then the point $(c,f(c))$ is the point of inflection.

Question 11. Find the number of critical points of the functions $$y = 2x + 3{x^{\frac{2}{3}}}$$

Solution: To find the number of critical points, let us differentiate the function w.r.t $x$ i.e.,

$$\begin{equation} \begin{aligned} f(x) = 2x + 3{x^{\frac{2}{3}}} \\ f'(x) = 2 + 3 \times \frac{2}{3}{x^{\frac{2}{3} - 1}} \\ f'(x) = 2 + 2{x^{ - \frac{1}{3}}} \\ f'(x) = 2 + \frac{2}{{{x^{\frac{1}{3}}}}} \\\end{aligned} \end{equation} $$ Case 1. When $f'(x)=0$, we get $$\begin{equation} \begin{aligned} f'(x) = 2 + \frac{2}{{{x^{\frac{1}{3}}}}} = 0 \\ {x^{\frac{1}{3}}} = - 1 \\ \Rightarrow x = - 1 \\\end{aligned} \end{equation} $$ Case 2. When $f'(x)$ is not defined i.e., $f'(x) = \infty $, we get $$\begin{equation} \begin{aligned} f'(x) = 2 + \frac{2}{{{x^{\frac{1}{3}}}}} = \infty \\ x = 0 \\\end{aligned} \end{equation} $$ Therefore, the function has two critical points i.e., $x=-1$ and $x=0$.

Question 12. Find the inflection point of $f(x) = 3{x^4} - 4{x^3}$. Also draw the graph of the function $f(x)$.

Solution: To find the inflection point, let us differentiate the function, we get $$\begin{equation} \begin{aligned} f(x) = 3{x^4} - 4{x^3} \\ f'(x) = 12{x^3} - 12{x^2} \\ f'(x) = 12{x^2}(x - 1) = 0 \\ \Rightarrow x = 0{\text{ and }}x = 1 \\\end{aligned} \end{equation} $$ Therefore, we get the critical points, i.e., $x = 0{\text{ and }}x = 1$. Now, we have to check the inflection points. We again differentiate it, we get $$\begin{equation} \begin{aligned} f'(x) = 12{x^3} - 12{x^2} \\ f''(x) = 36{x^2} - 24x \\ f''(x) = 12x(3x - 2) = 0 \\ \Rightarrow x = 0{\text{ and }}x = \frac{2}{3} \\\end{aligned} \end{equation} $$ Therefore, we have to check the sign of $f''(x)$ at $x = 0{\text{ and }}x = \frac{2}{3}$. Using number line method, we get $$\begin{equation} \begin{aligned} f''(x) > 0{\text{ when }}x < 0{\text{ and }}x > \frac{2}{3} \\ f''(x) < 0{\text{ when }}0 < x < \frac{2}{3} \\\end{aligned} \end{equation} $$ Therefore, $f''(x)$ changes sign at $x=0$ and $x = \frac{2}{3}$, therefore both are the inflection points. Now, to plot the graph of $f(x)$, mark the critical points and inflection points on the $X$-axis as shown in figure. We know that at $x=1$ there is a horizontal tangent, so draw the graph according to that. Also considering the sign of $f''(x)$, join the points $x=0$ and $x = \frac{2}{3}$ using the concavity and convexity as shown in figure.