I need help in the two questions
1 answer:
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All of given options contain quadratic functions. One way to determine the extreme value is squaring the expression with variable x.
Option B contain the expression where you can see perfect square. Thus, equation (choice B) reveals its extreme value without needing to be altered.
To determine the extreme value of this equation, you should substitute x=2 (x-value that makes expression in brackets equal to zero) into the function notation:
The extreme value of this equation has a minimum at the point (2,5).
First we find the zeroes so we don't take the integral of negative bits
4x-x²
x(4-x)
zeroes at x=0 and x=4
it opens down
so the part we are interested in is the bit between x=0 and x=4
![\int\limits^4_0 {4x-x^2} \, dx =[2x^2- \frac{1}{3}x^3]^4_0=(32- \frac{64}{3})-(0)= 10.6666666666](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E4_0%20%7B4x-x%5E2%7D%20%5C%2C%20dx%20%3D%5B2x%5E2-%20%5Cfrac%7B1%7D%7B3%7Dx%5E3%5D%5E4_0%3D%2832-%20%5Cfrac%7B64%7D%7B3%7D%29-%280%29%3D%20%2010.6666666666)
or aout 10 and 2/3
C is answer
4x-x²
x(4-x)
zeroes at x=0 and x=4
it opens down
so the part we are interested in is the bit between x=0 and x=4
C is answer