Question Statement

Use the second derivative to locate all points of inflection:

(i) $f(x)=x^4-x^3+2x^2+x-1$

(ii) $f(x)=x^{\frac{5}{3}}+4x$

(iii) $f(x)=\sin x$

(iv) $f(x)=\cos x$

(v) $f(x)=x-\sin x$

(vi) $f(x)=\tan x$

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Background and Explanation

Points of inflection occur where the concavity of a function changes, which requires the second derivative to equal zero or be undefined (with a sign change in $f^{\prime \prime }(x)$ around that point). To locate them, compute $f^{\prime \prime }(x)$, solve $f^{\prime \prime }(x)=0$, check for undefined points, and verify the concavity changes or simply identify the coordinates.

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Solution

Part (i): $f(x)=x^4-x^3+2x^2+x-1$

Given $f(x)=x^4-x^3+2x^2+x-1$

Differentiate w.r.t. $x$: $f^{\prime }(x)=4x^3-3x^2+4x+1$

Again differentiate w.r.t. $x$: $f^{\prime \prime }(x)=12x^2-6x+4$

For points of inflection, put $f^{\prime \prime }(x)=0$:

$\begin{array}{r}12x^2-6x+4=0\\ 2\left(6x^2-3x+2\right)=0\\ 6x^2-3x+2=0\end{array}$

Using the quadratic formula:

$\begin{array}{rl}x& =\frac{3\pm \sqrt{(-3{)}^2-4(6)(2)}}{2\times 6}\\ & =\frac{3\pm \sqrt{9-48}}{12}\\ x& =\frac{3\pm \sqrt{-39}}{12}\end{array}$

Since $\sqrt{-39}$ is imaginary, there are no real solutions for $x$. Hence there is no point of inflection.

Part (ii): $f(x)=x^{\frac{5}{3}}+4x$

Given $f(x)=x^{\frac{5}{3}}+4x$

Differentiate w.r.t. $x$:

$\begin{array}{rl}{f}^{\prime }(x)& =\frac{5}{3}{x}^{\frac{5}{3}-1}+4\\ f^{\prime }(x)& =\frac{5}{3}{x}^{\frac{2}{3}}+4\end{array}$

Again differentiate w.r.t. $x$:

$\begin{array}{rl}{f}^{\prime \prime }(x)& =\frac{5}{3}\cdot \frac{2}{3}{x}^{\frac{2}{3}-1}+0\\ & =\frac{10}{9}{x}^{-\frac{1}{3}}\\ f^{\prime \prime }(x)& =\frac{10}{9x^{\frac{1}{3}}}\end{array}$

For point of inflection, put $f^{\prime \prime }(x)=0$:

$\begin{array}{rl}\frac{10}{9x^{\frac{1}{3}}}& =0\\ 10& =0\end{array}$

which is not true.

Also $f^{\prime \prime }(x)$ is not defined at $x=0$. Put $x=0$ in eq. (1):

$f(0)=0+0=0$

Hence point of inflection is $(0,0)$.

Part (iii): $f(x)=\sin x$

Given $f(x)=\sin x$

Differentiate w.r.t. $x$: $f^{\prime }(x)=\cos x$

Again differentiate w.r.t. $x$: $f^{\prime \prime }(x)=-\sin x$

For points of inflection, put $f^{\prime \prime }(x)=0$:

$\begin{array}{rl}-\sin x& =0\\ \sin x& =0\\ \Rightarrow x& =n\pi ;n\in \mathbb{Z}\end{array}$

Hence points of inflection occur at $x=n\pi$. Put $x=n\pi$ in (1):

$f(n\pi )=\sin n\pi =0$

$\therefore$ Points of inflection are $(n\pi ,0)$ where $n\in \mathbb{Z}$.

Part (iv): $f(x)=\cos x$

Given $f(x)=\cos x$

Differentiate w.r.t. $x$: $f^{\prime }(x)=-\sin x$

Again differentiate w.r.t. $x$: $f^{\prime \prime }(x)=-\cos x$

For points of inflection, put $f^{\prime \prime }(x)=0$:

$\begin{array}{rl}-\cos x& =0\\ \cos x& =0\\ x& =(2n+1)\frac{\pi }{2};n\in \mathbb{Z}\end{array}$

Hence points of inflection occur at $x=(2n+1)\frac{\pi }{2}$. Put $x=(2n+1)\frac{\pi }{2}$ in (1):

$\begin{array}{rl}f\left((2n+1)\frac{\pi }{2}\right)& =\cos \left((2n+1)\frac{\pi }{2}\right)\\ & =0\end{array}$

$\therefore$ Points of inflection are $\left((2n+1)\frac{\pi }{2},0\right)$ where $n\in \mathbb{Z}$.

Part (v): $f(x)=x-\sin x$

Given $f(x)=x-\sin x$

Differentiate w.r.t. $x$: $f^{\prime }(x)=1-\cos x$

Again differentiate w.r.t. $x$:

$\begin{array}{rl}{f}^{\prime \prime }(x)& =0-(-\sin x)\\ f^{\prime \prime }(x)& =\sin x\end{array}$

For point of inflection, put $f^{\prime \prime }(x)=0$:

$\Rightarrow \begin{array}{rl}\sin x& =0\\ x& =n\pi ;n\in \mathbb{Z}\end{array}$

Put $x=n\pi$ in (1):

$\begin{array}{rl}f(n\pi )& =n\pi +\sin n\pi \\ & =n\pi +0\\ & =n\pi \end{array}$

Hence points of inflection are $(n\pi ,n\pi )$ where $n\in \mathbb{Z}$.

Part (vi): $f(x)=\tan x$

Given $f(x)=\tan x$

Differentiate w.r.t. $x$: $f^{\prime }(x)={\sec }^{2}x$

Again differentiate w.r.t. $x$:

$\begin{array}{rl}{f}^{\prime \prime }(x)& =2\sec x(\sec x\tan x)\\ & =2{\sec }^{2}x\tan x\end{array}$

For point of inflection, put $f^{\prime \prime }(x)=0$:

$\begin{array}{rl}2{\sec }^{2}x\tan x& =0\end{array}$

This implies either $2{\sec }^{2}x=0$ or $\tan x=0$.

$\begin{array}{rl}{\sec }^{2}x& =0\end{array}$

which is not possible because ${\sec }^{2}x$ can never be zero for any $x$.

Therefore: $\begin{array}{rl}\tan x& =0\\ x& =n\pi ;n\in \mathbb{Z}\end{array}$

Put $x=n\pi$ in (1):

$f(n\pi )=\tan n\pi =0$

$\therefore$ Points of inflection are $(n\pi ,0)$ where $n\in \mathbb{Z}$.

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Key Formulas or Methods Used

• Second derivative test for inflection points: Set $f^{\prime \prime }(x)=0$ or identify where $f^{\prime \prime }(x)$ is undefined, then check for concavity changes
• Power rule for differentiation: $\frac{d}{dx}(x^n)=n{x}^{n-1}$
• Trigonometric derivatives: $\frac{d}{dx}(\sin x)=\cos x$, $\frac{d}{dx}(\cos x)=-\sin x$, $\frac{d}{dx}(\tan x)={\sec }^{2}x$, $\frac{d}{dx}(\sec x)=\sec x\tan x$
• Quadratic formula: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ for solving $a{x}^2+bx+c=0$
• General solutions for trigonometric equations: $\sin x=0\Rightarrow x=n\pi$ and $\cos x=0\Rightarrow x=(2n+1)\frac{\pi }{2}$ where $n\in \mathbb{Z}$

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Summary of Steps

1. Differentiate the given function $f(x)$ to find the first derivative $f^{\prime }(x)$.
2. Differentiate again to obtain the second derivative $f^{\prime \prime }(x)$.
3. Solve $f^{\prime \prime }(x)=0$ to find potential inflection points (check where $f^{\prime \prime }(x)$ is undefined as well).
4. Analyze the solutions: if the equation has no real solutions, there are no inflection points; if solutions exist, substitute the $x$-values back into the original function $f(x)$.
5. State the coordinates of all points of inflection as $(x,f(x))$, including the general form for periodic functions (e.g., $n\in \mathbb{Z}$).