Answer to Question #160294 in Calculus for sdfad

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\n1.\\\\\n(i)\\\\\nf(x) = \\sqrt[3]{x^2} = x^{\\frac{2}{3}} \\\\\n\nf'(x) = \\frac{2}{3} x^{-\\frac{1}{3}} \\\\\n\n\n\\textsf{At stationary point},\\,\\, f'(x) = 0 \\\\\n\n\\frac{2}{3} x^{-\\frac{1}{3}}= 0,\\,\\, \\frac{2}{3} x^{-\\frac{1}{3}} \\times x = 0 \\times x \\\\\n\n\\frac{2}{3} x^{\\frac{2}{3}}= 0\n\n\\therefore x = 0\n\n\n\\textsf{Absolute maximum:} \\,\\,f(1) = f(-1) = \\sqrt[3]{1^2} = 1 \\\\\n\n\\textsf{Absolute minimum:} \\,\\,f(0) = \\sqrt[3]{0^2} = 0 \\\\\n\n(ii)\\\\ g(x) = xe^{2x} \\\\\n\ng'(x) = e^{2x} + 2xe^{2x} = e^{2x}(1 + 2x) \\\\\n\n\n\\textsf{At stationary point,}\\,\\,g'(x) = 0 \\\\\n\ne^{2x}(1 + 2x) = 0,\\,\\, x = -\\frac{1}{2} \\\\\n\n\n\\textsf{Absolute maximum:}\\,\\, g(0) = 0 \\\\\n\n\\textsf{Absolute minimum:}\\,\\, g\\left(-\\frac{1}{2}\\right) = -\\frac{1}{2e} \\\\\n\n2\\\\\n(i)\\\\\nf(x) = 3x^3 - 18x^2 - 45x + 10 \\\\\n\nf'(x) = 9x^2 - 36x - 45 \\\\\n\nf(x)\\,\\, \\textsf{is increasing if}\\,\\, f'(x) > 0 \\\\\n\n9x^2 - 36x - 45 > 0 \\\\\n\nx^2 - 4x - 5 > 0 \\\\\n\n(x + 1)(x - 5) > 0 \\\\\n\nx < -1,\\,\\, x > 5. \\\\\n\n(ii)\\\\\n\\textsf{At stationary point,}\\,\\, f'(x) = 0 \\\\\n\n\\therefore x = -1, 5 \\\\\n\n\nf''(x) = 18x - 36,\\,\\, f''(-1) = -54 < 0 \\\\\n\nf''(5) = 18(5) - 36 = 90 - 36 = 54 > 0\\\\\n\n\n\\therefore \\textsf{By the second derivative test,}\\\\\n\\textsf{at}\\,\\, x = 5\\,\\, \\textsf{there is a minimum.} \\\\\n\n\n(iii)\\\\ \nf(x)\\,\\, \\textsf{is decreasing and concave up}\\,\\,\\textsf{if}\\,\\, f'(x) < 0. \\\\\n\n\\therefore -1 < x < 5\\\\\n\nf''(x) = 18x - 36,\\\\\n\nf(x)\\,\\, \\textsf{is concave up}\\,\\, \\textsf{if}\\,\\,f"(x) > 0\\\\\n\n18x - 36 > 0 \\\\\n18x > 36\\\\\nx > 2\\\\\n\n\n\\therefore f(x)\\,\\, \\textsf{is concave up}\\,\\, \\textsf{for}\\,\\, x > 2.\\\\\n\n\\textsf{The graph is decreasing at that point}\\\\\n\\textsf{since it lies between}\\,\\, 2 \\,\\, \\textsf{and}\\,\\, 5.\\\\\n\n\n\n\\therefore \\textsf{We can conclude that}\\\\\nf(x)\\,\\, \\textsf{is concave up and decreasing}\\,\\, \\textsf{for}\\,\\, x > 2.\\\\"