consider the following. (if an answer does not exist, enter dne.)\n\n$f(x)=\\frac{x^{2}-15}{x - 4}$\n\n(a)…

consider the following. (if an answer does not exist, enter dne.)\n\n$f(x)=\\frac{x^{2}-15}{x - 4}$\n\n(a) find the interval(s) on which $f$ is increasing. (enter your answer using interval notation.)\n\n(b) find the interval(s) on which $f$ is decreasing. (enter your answer using interval notation.)\n\n(c) find the local minimum and maximum value of $f$.\n\nlocal minimum value\n\nlocal maximum value

consider the following. (if an answer does not exist, enter dne.)\n\n$f(x)=\\frac{x^{2}-15}{x - 4}$\n\n(a) find the interval(s) on which $f$ is increasing. (enter your answer using interval notation.)\n\n(b) find the interval(s) on which $f$ is decreasing. (enter your answer using interval notation.)\n\n(c) find the local minimum and maximum value of $f$.\n\nlocal minimum value\n\nlocal maximum value

Answer

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). [ \begin{align*} f^\prime(x)&=\frac{(2x)(x - 4)-(x^{2}-15)(1)}{(x - 4)^{2}}\ &=\frac{2x^{2}-8x-x^{2}+15}{(x - 4)^{2}}\ &=\frac{x^{2}-8x + 15}{(x - 4)^{2}}\ &=\frac{(x-3)(x - 5)}{(x - 4)^{2}} \end{align*} ] The domain of (f(x)) is (x\neq4).

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).