Question Statement

Find the critical values of the following functions:

(i) $f(x)=2x^2-6x+8$ (ii) $f(x)=x^3+x-2$ (iii) $f(x)=\frac{x}{x^2+2}$ (iv) $f(x)=\cos 4x$ (v) $f(x)=(4x-3{)}^{\frac{1}{3}}$ (vi) $f(x)=x^2(x+1{)}^3$

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Background and Explanation

A critical value of a function $f(x)$ is a value $x$ in the domain of the function where the first derivative is either zero ($f^{\prime }(x)=0$) or undefined. These points are essential in calculus for finding local maxima, minima, and points of inflection.

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Solution

Part (i)

Given: $f(x)=2x^2-6x+8$

Step 1: Differentiate with respect to $x$ $\begin{array}{rl}{f}^{\prime }(x)& =2\frac{d}{dx}\left(x^2\right)-6\frac{d}{dx}(x)+\frac{d}{dx}(8)\\ & =2(2x)-6(1)+0\\ f^{\prime }(x)& =4x-6\end{array}$

Step 2: Set the derivative to zero to find critical values $\begin{array}{rl}4x-6& =0\\ 4x& =6\\ x& =\frac{6}{4}\\ x& =\frac{3}{2}\end{array}$ Conclusion: The critical value is $x=\frac{3}{2}$.

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Part (ii)

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

Step 1: Differentiate with respect to $x$ $\begin{array}{rl}{f}^{\prime }(x)& =\frac{d}{dx}\left(x^3\right)+\frac{d}{dx}(x)-\frac{d}{dx}(2)\\ & =3x^2+1-0\\ f^{\prime }(x)& =3x^2+1\end{array}$

Step 2: Set the derivative to zero $\begin{array}{rl}3x^2+1& =0\\ 3x^2& =-1\\ x^2& =\frac{-1}{3}\\ x& =\pm \frac{1}{\sqrt{3}}i\end{array}$ Conclusion: Since the values are imaginary, there are no real critical values for this function.

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Part (iii)

Given: $f(x)=\frac{x}{x^2+2}$

Step 1: Differentiate using the Quotient Rule $\begin{array}{rl}{f}^{\prime }(x)& =\frac{\left(x^2+2\right)\frac{d}{dx}(x)-x\frac{d}{dx}\left(x^2+2\right)}{{\left(x^2+2\right)}^2}\\ & =\frac{\left(x^2+2\right)\cdot 1-x(2x+0)}{{\left(x^2+2\right)}^2}\\ & =\frac{x^2+2-2x^2}{{\left(x^2+2\right)}^2}\\ f^{\prime }(x)& =\frac{2-x^2}{{\left(x^2+2\right)}^2}\end{array}$

Step 2: Set the derivative to zero $\frac{2-x^2}{{\left(x^2+2\right)}^2}=0$ $\begin{array}{rl}2-x^2& =0\\ x^2& =2\\ x& =\pm \sqrt{2}\end{array}$ Conclusion: The critical values are $x=\pm \sqrt{2}$.

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Part (iv)

Given: $f(x)=\cos 4x$

Step 1: Differentiate using the Chain Rule $\begin{array}{rl}{f}^{\prime }(x)& =\frac{d}{dx}(\cos 4x)\\ & =-\sin 4x\cdot \frac{d}{dx}(4x)\\ & =-\sin 4x\cdot 4\\ f^{\prime }(x)& =-4\sin 4x\end{array}$

Step 2: Set the derivative to zero $\begin{array}{rl}-4\sin 4x& =0\\ \sin 4x& =0\\ 4x& ={\sin }^{-1}(0)\\ 4x& =n\pi (\text{where }n\in \mathbb{Z})\\ x& =\frac{n\pi }{4};n\in \mathbb{Z}\end{array}$ Conclusion: The critical values are $x=\frac{n\pi }{4}$ for any integer $n$.

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Part (v)

Given: $f(x)=(4x-3{)}^{\frac{1}{3}}$

Step 1: Differentiate using the Power Rule and Chain Rule $\begin{array}{rl}{f}^{\prime }(x)& =\frac{1}{3}(4x-3{)}^{\frac{1}{3}-1}\cdot \frac{d}{dx}(4x-3)\\ & =\frac{1}{3}(4x-3{)}^{-\frac{2}{3}}\cdot (4)\\ f^{\prime }(x)& =\frac{4}{3(4x-3{)}^{2/3}}\end{array}$

Step 2: Check where the derivative is zero or undefined Setting $f^{\prime }(x)=0$: $\frac{4}{3(4x-3{)}^{2/3}}=0⟹4=0(\text{Not possible})$ Setting the denominator to zero (where $f^{\prime }(x)$ is undefined): $\begin{array}{rl}3(4x-3{)}^{2/3}& =0\\ 4x-3& =0\\ 4x& =3\\ x& =\frac{3}{4}\end{array}$ Conclusion: The critical value is $x=\frac{3}{4}$ (where the derivative is undefined).

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Part (vi)

Given: $f(x)=x^2(x+1{)}^3$

Step 1: Differentiate using the Product Rule and Chain Rule $\begin{array}{rl}{f}^{\prime }(x)& =\frac{d}{dx}(x^2)\cdot (x+1{)}^3+x^2\cdot \frac{d}{dx}(x+1{)}^3\\ & =2x(x+1{)}^3+x^2\cdot 3(x+1{)}^2\cdot (1)\\ f^{\prime }(x)& =2x(x+1{)}^3+3x^2(x+1{)}^2\end{array}$

Step 2: Factor the derivative and set to zero $\begin{array}{rl}x(x+1{)}^2[2(x+1)+3x]& =0\\ x(x+1{)}^2(2x+2+3x)& =0\\ x(x+1{)}^2(5x+2)& =0\end{array}$

Solving for $x$:

1. $x=0$
2. $(x+1{)}^2=0⟹x=-1$
3. $5x+2=0⟹x=-\frac{2}{5}$

Conclusion: The critical values are $x=0,-1,-\frac{2}{5}$.

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Key Formulas or Methods Used

• Power Rule: $\frac{d}{dx}(x^n)=n{x}^{n-1}$
• Product Rule: $\frac{d}{dx}[u(x)v(x)]=u^{\prime }v+u{v}^{\prime }$
• Quotient Rule: $\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right]=\frac{v{u}^{\prime }-u{v}^{\prime }}{v^2}$
• Chain Rule: $\frac{d}{dx}[f(g(x))]=f^{\prime }(g(x))\cdot g^{\prime }(x)$
• Trigonometric Derivative: $\frac{d}{dx}(\cos x)=-\sin x$
• General Solution: $\sin \theta =0⟹\theta =n\pi$

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Summary of Steps

1. Find the first derivative $f^{\prime }(x)$ of the given function.
2. Set $f^{\prime }(x)=0$ and solve for $x$.
3. Identify any values of $x$ where $f^{\prime }(x)$ is undefined (but $f(x)$ is defined).
4. The resulting $x$ values are the critical values.