Question

consider the following. (if an answer does not exist, enter dne.)

$f(x)=\frac{x^{2}-15}{x - 4}$

(a) find the interval(s) on which $f$ is increasing. (enter your answer using interval notation.)

(b) find the interval(s) on which $f$ is decreasing. (enter your answer using interval notation.)

(c) find the local minimum and maximum value of $f$.

local minimum value

local maximum value

Explanation:

Step1: Find the derivative of \(f(x)\)

Use the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\), where \(u = x^{2}-15\), \(u^\prime=2x\), \(v=x - 4\), \(v^\prime = 1\).

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The domain of \(f(x)\) is \(x
eq4\).

Step2: Find the critical points

Set \(f^\prime(x)=0\), then \((x - 3)(x - 5)=0\), so \(x=3\) or \(x = 5\).

Step3: Test the intervals

• For the interval \((-\infty,3)\), let \(x=2\), then \(f^\prime(2)=\frac{(2 - 3)(2 - 5)}{(2 - 4)^{2}}=\frac{(-1)\times(-3)}{4}=\frac{3}{4}>0\).
• For the interval \((3,4)\), let \(x=\frac{7}{2}\), then \(f^\prime(\frac{7}{2})=\frac{(\frac{7}{2}-3)(\frac{7}{2}-5)}{(\frac{7}{2}-4)^{2}}=\frac{(\frac{1}{2})(-\frac{3}{2})}{(\frac{-1}{2})^{2}}=- 3<0\).
• For the interval \((4,5)\), let \(x=\frac{9}{2}\), then \(f^\prime(\frac{9}{2})=\frac{(\frac{9}{2}-3)(\frac{9}{2}-5)}{(\frac{9}{2}-4)^{2}}=\frac{(\frac{3}{2})(-\frac{1}{2})}{(\frac{1}{2})^{2}}=-3<0\).
• For the interval \((5,\infty)\), let \(x=6\), then \(f^\prime(6)=\frac{(6 - 3)(6 - 5)}{(6 - 4)^{2}}=\frac{3\times1}{4}=\frac{3}{4}>0\).

Answer:

(a) The function \(f(x)\) is increasing on \((-\infty,3)\cup(5,\infty)\).
(b) The function \(f(x)\) is decreasing on \((3,4)\cup(4,5)\).
(c) Local minimum value: \(f(5)=\frac{5^{2}-15}{5 - 4}=10\); Local maximum value: \(f(3)=\frac{3^{2}-15}{3 - 4}=6\).