Answer to Question #177792 in Calculus for John green

Question #177792

For the function f(x)=x3-6x2-12 determine


a) intervals of increase and decrease


b) location of max and minimum points


c) interval of concavity down or up


d) locator of points of inflection


1
Expert's answer
2021-04-15T07:21:12-0400

To solve the items find previously first and second derivatives of given function

"f(x)=x^3-6x^2-12"


"f'(x)=3x^2-6\\cdot2x-0=3x^2-12x"

"f''(x)=3\\cdot2x-12=6x-12"


a) intervals of increase and decrease

Find intervals of increase solving the inequality

"f'(x)>0"

"3x^2-12x>0"

"3x(x-4)>0"

"\\left\\{\\begin{array}{l}3>0\\\\ x(x-4)>0\\end{array}\\right.\\Rightarrow" "x(x-4)>0\\Rightarrow"


"\\left[\\begin{array}{l}\\left\\{\\begin{array}{l}x>0\\\\x-4>0\\end{array}\\right.\\\\ \\\\ \\left\\{\\begin{array}{l}x<0\\\\x-4<0\\end{array}\\right.\\end{array}\\right.\\Rightarrow" "\\left[\\begin{array}{l}\\left\\{\\begin{array}{l}x>0\\\\x>4\\end{array}\\right.\\\\ \\\\ \\left\\{\\begin{array}{l}x<0\\\\x<4\\end{array}\\right.\\end{array}\\right.\\Rightarrow" "\\left[\\begin{array}{l}x>4 \\\\ x<0\\end{array}\\right.\\Rightarrow" "x\\in(-\\infty;0)\\cup(4;+\\infty)"

Find intervals of decrease solving the inequality

"f'(x)<0"

"3x^2-12x<0"

"3x(x-4)<0"

"\\left\\{\\begin{array}{l}3>0\\\\ x(x-4)<0\\end{array}\\right.\\Rightarrow" "x(x-4)<0\\Rightarrow"


"\\left[\\begin{array}{l}\\left\\{\\begin{array}{l}x>0\\\\x-4<0\\end{array}\\right.\\\\ \\\\ \\left\\{\\begin{array}{l}x<0\\\\x-4>0\\end{array}\\right.\\end{array}\\right.\\Rightarrow" "\\left[\\begin{array}{l}\\left\\{\\begin{array}{l}x>0\\\\x<4\\end{array}\\right.\\\\ \\\\ \\left\\{\\begin{array}{l}x<0\\\\x>4\\end{array}\\right.\\end{array}\\right.\\Rightarrow" "\\left[\\begin{array}{l}0<x<4 \\\\ x\\in\\emptyset\\end{array}\\right.\\Rightarrow" "x\\in(0;4)"



b) location of maximum and minimum points

Find critical points solving the equation

"f'(x)=0"

"3x^2-12x=0"

"3x(x-4)=0"

"\\left[\\begin{array}{l}x=0 \\\\ x=4\\end{array}\\right."

At the point "x_1=0" the derivative swithes its sign from "+" to "-" . Hence the function switches from increasing to decreasing at the point. Thus "x_1=0" is the maximum point.

"f(0)=12"

At the point "x_2=4" the derivative swithes its sign from "-" to "+" . Hence the function switches from decreasing to increasing at the point. Thus "x_2=4" is the minimum point.

"f(4)=4^3-6\\cdot4^2-12=64-96-12=-44"


c) intervals of concavity down or up

Find intervals of concavity down solving the inequality

"f''(x)>0"

"6x-12>0"

"6(x-2)>0"

"\\left\\{\\begin{array}{l}6>0\\\\ x-2>0\\end{array}\\right.\\Rightarrow" "x>2\\Rightarrow" "x\\in(2;+\\infty)"


Find intervals of concavity up solving the inequality

"f''(x)<0"

"6x-12<0"

"6(x-2)<0"

"\\left\\{\\begin{array}{l}6>0\\\\ x-2<0\\end{array}\\right.\\Rightarrow" "x<2\\Rightarrow" "x\\in(-\\infty;2)"


d) locator of points of inflection

Find points of inflection solving the equation

"f''(x)=0"

"6x-12=0"

"6(x-2)=0"

"x=2"

At the point "x=2" the second derivative swithes its sign from "-" to "+" . Hence the graph of function switches from concavity up to concavity down at the point. Thus "x=2" is the inflection point.

"f(2)=2^3-6\\cdot2^2-12=8-24-12=-28"



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