Answer to Question #250090 in Electrical Engineering for Nazi

Question #250090
1.find the absolute maximum and minimum of the function f(x)=4X^2-7X+3 on the interval [-2,3]
2. find the absolute maximum and minimum of f(x)=4X^3-8X^2+1 on the closed interval [-1,1]
3. find the absolute maximum and minimum of f(x)=X^3/(x+2) in the interval [-1,1]
1
Expert's answer
2021-10-13T04:57:34-0400

"1)\\\\f(x)=4x^2-7x+3\\\\\nf'(x)=8x-7\\\\\nf'(x)=0=8x-7\\\\\nx=\\frac{7}{8}\\\\\n\\text{f(x) at }\\frac{7}{8}\\text{ is }\\frac{-1}{16}\\\\\n\\text{f(x) at -2 and 3 is 33 and 18 respectively}\\\\ \n\\therefore \\text{ absolute max occur at (-2,33)}\\\\\n\\text{Absolute min occur at }(\\frac{7}{8},\\frac{-1}{16})\\\\\n2)\\\\\nf(x)=4x^3-8x^2+1\\\\\nf'(x)=12x^2-16x\\\\\nx=0,\\frac{4}{3}\\\\\nf(0)=1\\\\\nf(\\frac{4}{3} )=\\frac{-101}{27}\\\\\nf(-1)=-11\\\\\nf(1)=-3\\\\\n\\text{absolute max is (0,1) while absolute min is (-1,-11)}\\\\\n3)\\\\\nf(x)=\\frac{x^3}{x+2}\\\\\nf'(x)=\\frac{3x^2(x+2)-x^3}{(x+2)^2}\\\\\nf'(x)=0\\\\\nx^2-3x-8=0\\\\\nx=4.37,-1.37\\\\\nf(4.37)=13.1\\\\\nf(1.37)=0.76\\\\\nf(-1)=-1\\\\\nf(1)=\\frac{1}{3}\\\\\n\\text{Absolute max is (4.37,13.1) and absolute min is (-1,-1)}"


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