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)=4x27x+3f(x)=8x7f(x)=0=8x7x=78f(x) at 78 is 116f(x) at -2 and 3 is 33 and 18 respectively absolute max occur at (-2,33)Absolute min occur at (78,116)2)f(x)=4x38x2+1f(x)=12x216xx=0,43f(0)=1f(43)=10127f(1)=11f(1)=3absolute max is (0,1) while absolute min is (-1,-11)3)f(x)=x3x+2f(x)=3x2(x+2)x3(x+2)2f(x)=0x23x8=0x=4.37,1.37f(4.37)=13.1f(1.37)=0.76f(1)=1f(1)=13Absolute max is (4.37,13.1) and absolute min is (-1,-1)1)\\f(x)=4x^2-7x+3\\ f'(x)=8x-7\\ f'(x)=0=8x-7\\ x=\frac{7}{8}\\ \text{f(x) at }\frac{7}{8}\text{ is }\frac{-1}{16}\\ \text{f(x) at -2 and 3 is 33 and 18 respectively}\\ \therefore \text{ absolute max occur at (-2,33)}\\ \text{Absolute min occur at }(\frac{7}{8},\frac{-1}{16})\\ 2)\\ f(x)=4x^3-8x^2+1\\ f'(x)=12x^2-16x\\ x=0,\frac{4}{3}\\ f(0)=1\\ f(\frac{4}{3} )=\frac{-101}{27}\\ f(-1)=-11\\ f(1)=-3\\ \text{absolute max is (0,1) while absolute min is (-1,-11)}\\ 3)\\ f(x)=\frac{x^3}{x+2}\\ f'(x)=\frac{3x^2(x+2)-x^3}{(x+2)^2}\\ f'(x)=0\\ x^2-3x-8=0\\ x=4.37,-1.37\\ f(4.37)=13.1\\ f(1.37)=0.76\\ f(-1)=-1\\ f(1)=\frac{1}{3}\\ \text{Absolute max is (4.37,13.1) and absolute min is (-1,-1)}


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS