Answer to Question #160294 in Calculus for sdfad

Question #160294

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

1) 𝑓(π‘₯) = Cube root of x^2 on the interval [βˆ’1,1]Β 

2) 𝑔(π‘₯) = π‘₯𝑒^2π‘₯ on the interval [βˆ’2,0]Β 


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

Β Given 𝑓(π‘₯) = 3π‘₯ ^3 βˆ’ 18π‘₯^ 2 βˆ’ 45π‘₯ + 10

a. On what interval(s), is 𝑓(π‘₯) increasing? Show all work/justify


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



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


1
Expert's answer
2021-02-04T06:19:07-0500

"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.\\\\"


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