Question

Please help with ALL parts of #13
1 = 13) Find absolute extrema for f(x) on the interval [0, 1]. Is the Extreme value Theorem satisfied? If not, use graphing calculator to find the absolute extrema if any. x-x2

Answer 1

13. Consider the function
f(x) = 1/(x - x^2) = (x - x^2)^-1
now, in the interval [0,1]
f(0) = inf, f(1) = inf
So inside the interval [0,1] the extreme value (maximum) reached by the function is inf, this means there is a minima in between
now, from symmetry at f(0.5) = 1/(0.5 - 0.25) = 4
hence, minimum value of the function f(x) in interval [0,1] is 4
and yes, extreme value theorem is satisfeid as the function is continuous between x = 0 and x = 1 and hence there should exist a minimum and maximum value for the said function in shte said interval
The graph fro the function is as under

Graph for 1/(x-x^2) x: 1.14081126 y: -6.22513543 14+ 12 10 8 6 4 -0.3 -0.2 -0.1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.1 1.2 More info

15. function f(x) = sin(x) + cos(x)
f(x) = sqrt(2)[sin(x)/sqrt(2) + cos(x)/sqrt(2)]
f(x) = sqrt(2)(sin(x)cos(45) + cos(x)sin(45))
f(x) = sqrt(2)*sin(x + 45)
hence, in the interval [0, pi]
consider f'(x) = 0
then
cos(x + 45) = 0
hence x = pi/4
hence there is just one extremum in between the points [0,pi]
so f(0) = sqrt(2)*cos(45) = 1
f(pi) = sqrt(2)*sin(pi + pi/4) = -1
f(pi/4) = sqrt(2)*sin(pi/2) = sqrt(2)
so the maximum of the fucntion in interval [0,1] is sqrt(2) with minima being -1