Question

Locate the absolute maximum and minimum for each of the following functions and justify your responses. show all work!

1) 𝑓(𝑥) = Cube root of x^2 on the interval [−1,1] 

2) 𝑔(𝑥) = 𝑥𝑒^2𝑥 on the interval [−2,0] 

For each of the following functions, respond to the given prompts. Show all work that leads to your responses. Show all work!

 Given 𝑓(𝑥) = 3𝑥 ^3 − 18𝑥^ 2 − 45𝑥 + 10

a. On what interval(s), is 𝑓(𝑥) increasing? Show all work/justify

b. At what value(s) of 𝑥 does 𝑓(𝑥) have a relative minimum? show all work/justify

c. On what interval(s), is 𝑓(𝑥) decreasing and concave up? show all work/justify

Answer 1

$It\displaystyle 1.\\ (i)\\ f(x) = \sqrt[3]{x^2} = x^{\frac{2}{3}} \\ f'(x) = \frac{2}{3} x^{-\frac{1}{3}} \\ \textsf{At stationary point},\,\, f'(x) = 0 \\ \frac{2}{3} x^{-\frac{1}{3}}= 0,\,\, \frac{2}{3} x^{-\frac{1}{3}} \times x = 0 \times x \\ \frac{2}{3} x^{\frac{2}{3}}= 0 \therefore x = 0 \textsf{Absolute maximum:} \,\,f(1) = f(-1) = \sqrt[3]{1^2} = 1 \\ \textsf{Absolute minimum:} \,\,f(0) = \sqrt[3]{0^2} = 0 \\ (ii)\\ g(x) = xe^{2x} \\ g'(x) = e^{2x} + 2xe^{2x} = e^{2x}(1 + 2x) \\ \textsf{At stationary point,}\,\,g'(x) = 0 \\ e^{2x}(1 + 2x) = 0,\,\, x = -\frac{1}{2} \\ \textsf{Absolute maximum:}\,\, g(0) = 0 \\ \textsf{Absolute minimum:}\,\, g\left(-\frac{1}{2}\right) = -\frac{1}{2e} \\ 2\\ (i)\\ f(x) = 3x^3 - 18x^2 - 45x + 10 \\ f'(x) = 9x^2 - 36x - 45 \\ f(x)\,\, \textsf{is increasing if}\,\, f'(x) > 0 \\ 9x^2 - 36x - 45 > 0 \\ x^2 - 4x - 5 > 0 \\ (x + 1)(x - 5) > 0 \\ x < -1,\,\, x > 5. \\ (ii)\\ \textsf{At stationary point,}\,\, f'(x) = 0 \\ \therefore x = -1, 5 \\ f''(x) = 18x - 36,\,\, f''(-1) = -54 < 0 \\ f''(5) = 18(5) - 36 = 90 - 36 = 54 > 0\\ \therefore \textsf{By the second derivative test,}\\ \textsf{at}\,\, x = 5\,\, \textsf{there is a minimum.} \\ (iii)\\ f(x)\,\, \textsf{is decreasing and concave up}\,\,\textsf{if}\,\, f'(x) < 0. \\ \therefore -1 < x < 5\\ f''(x) = 18x - 36,\\ f(x)\,\, \textsf{is concave up}\,\, \textsf{if}\,\,f"(x) > 0\\ 18x - 36 > 0 \\ 18x > 36\\ x > 2\\ \therefore f(x)\,\, \textsf{is concave up}\,\, \textsf{for}\,\, x > 2.\\ \textsf{The graph is decreasing at that point}\\ \textsf{since it lies between}\,\, 2 \,\, \textsf{and}\,\, 5.\\ \therefore \textsf{We can conclude that}\\ f(x)\,\, \textsf{is concave up and decreasing}\,\, \textsf{for}\,\, x > 2.\\$