Question

1. (calculator not allowed) for what value of x does the function f(x)=(x - 2)(x - 3)^2 have a relative maximum? (a) -3 (b) -7/3 (c) -5/2 (d) 7/3 (e) 5/2 5. (calculator not allowed) what is the minimum value of f(x)=x ln x? (a) -e (b) -1 (c) -1/e (d) 0 (e) f(x) has no minimum value. 6. (calculator not allowed) the derivative of f is x^4(x - 2)(x + 3). at how many points will the graph of f have a relative maximum? (a) none (b) one (c) two (d) three (e) four

Explanation:

4.

Step1: Expand the function

\[

$$\begin{align*} f(x)&=(x - 2)(x - 3)^{2}\\ &=(x - 2)(x^{2}-6x + 9)\\ &=x^{3}-6x^{2}+9x-2x^{2}+12x - 18\\ &=x^{3}-8x^{2}+21x - 18 \end{align*}$$

\]

Step2: Find the derivative

Using the power - rule \((x^n)^\prime=nx^{n - 1}\), \(f^\prime(x)=3x^{2}-16x + 21\).

Step3: Set the derivative equal to zero

\[3x^{2}-16x + 21 = 0\]
The quadratic formula for \(ax^{2}+bx + c=0\) is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), here \(a = 3\), \(b=-16\), \(c = 21\).
\[x=\frac{16\pm\sqrt{(-16)^{2}-4\times3\times21}}{2\times3}=\frac{16\pm\sqrt{256 - 252}}{6}=\frac{16\pm2}{6}\]
The critical points are \(x=\frac{16 + 2}{6}=3\) and \(x=\frac{16-2}{6}=\frac{7}{3}\).

Step4: Use the second - derivative test

Find the second - derivative \(f^{\prime\prime}(x)=6x-16\).
\(f^{\prime\prime}(\frac{7}{3})=6\times\frac{7}{3}-16=14 - 16=-2<0\), so \(x = \frac{7}{3}\) is a relative maximum.
\(f^{\prime\prime}(3)=6\times3-16=18 - 16 = 2>0\), so \(x = 3\) is a relative minimum.

Step1: Find the domain

The function \(y = f(x)=x\ln x\) has domain \(x>0\).

Step2: Find the derivative

Using the product rule \((uv)^\prime=u^\prime v+uv^\prime\), where \(u = x\), \(v=\ln x\), \(u^\prime=1\), \(v^\prime=\frac{1}{x}\). So \(f^\prime(x)=\ln x+1\).

Step3: Set the derivative equal to zero

\(\ln x+1 = 0\), then \(\ln x=-1\).
By the definition of the natural logarithm, if \(\ln x=-1\), then \(x = e^{-1}=\frac{1}{e}\).

Step4: Use the second - derivative test

Find the second - derivative \(f^{\prime\prime}(x)=\frac{1}{x}\).
When \(x=\frac{1}{e}\), \(f^{\prime\prime}(\frac{1}{e})=e>0\), so \(x=\frac{1}{e}\) is a relative minimum.
Substitute \(x = \frac{1}{e}\) into \(f(x)\): \(f(\frac{1}{e})=\frac{1}{e}\ln\frac{1}{e}=-\frac{1}{e}\).

Step1: Find the critical points

Set \(f^\prime(x)=x^{4}(x - 2)(x + 3)=0\).
The critical points are \(x = 0\), \(x = 2\), and \(x=-3\).

Step2: Use the sign - change test for the derivative

Consider the intervals \((-\infty,-3)\), \((-3,0)\), \((0,2)\) and \((2,\infty)\).
For \(x<-3\), let \(x=-4\), then \(f^\prime(-4)=(-4)^{4}(-4 - 2)(-4 + 3)=256\times(-6)\times(-1)>0\).
For \(-3For \(0For \(x>2\), let \(x = 3\), then \(f^\prime(3)=3^{4}(3 - 2)(3 + 3)=81\times1\times6>0\).
The sign of \(f^\prime(x)\) changes from positive to negative only at \(x=-3\).

Answer:

D. \(\frac{7}{3}\)

5.