True or False: If f(x) is a strictly decreasing function defined for all values of x, then f(0) < f(1).?

If f is strictly decreasing, then the larger x is, the smaller f(x) is, so f(0)%26gt;f(1).





Therefore the answer is False.





Another way to see this is the proof through calculus (if f was differentialbe).





By the mean value theorem, there exists some c such that 0%26lt;c%26lt;1 and f'(c)=(f(1)-f(0))/(1-0) =f(1)-f(0). Since f is strictly decreasing f'(c)%26lt;0 (this is true for all x, not just c). Therefore f(1)-f(0)=f'(c)%26lt;0 or f(1)%26lt;f(0).True or False: If f(x) is a strictly decreasing function defined for all values of x, then f(0) %26lt; f(1).?
Since f(x) is decreasing as x increases. Then f(0) %26lt; f(1) is false.





f(x) = -x is an example of such a function.


f(0) = 0


f(1) = -1


0 %26lt; -1 is falseTrue or False: If f(x) is a strictly decreasing function defined for all values of x, then f(0) %26lt; f(1).?
f(0)is 0 and has no assigned value and f(1) is therefore %26gt; f(0).


The ans. then must be true.
False.





By definition of a decreasing function,





when x gets larger,





the f(x) gets smaller.





Given x1 and x2, such that x1 %26lt; x2 (note: x gets larger)





Then f(x1) %26gt; f(x2) (note: f(x) gets smaller)