<p>Question is irrenelvent to this forum but still bugging me. I need an example of a situation in which one function’s value is always greater than another function, for all real values of x, but its deriviative is always less than the derviative of the other function.</p>
<p>How about Y=e^x and Y=-1?</p>
<p>Deriv of first is e^x, for all real x, e^x>0, deriv of Y=-1 is 0…</p>
<p>Ya, but e^x is always greater than y=-1 and its derivitive is always greater. You need a function who is always greater than another function but whose derivitive is always less than the other function.</p>
<p>i thought …
f(x)= a positive constant for the one which has greater value but less deriv
and
g(x)= -x^2 for second since less value but greater deriv? I am unclear if a deriv of -2x would be greater than a deriv 0 (from first function) but it seems like it should since it has a greater change…</p>
<p>f(x)= -arctan(x) for first equation (arctan is inverse tangent). This function has a range of -pi/2<y<pi 2.=“” no=“” value=“” for=“” y=“” will=“” be=“” lower=“” than=“” -pi=“” g(x)=“-3” second=“” equation.=“” always=“” -3=“” here.=“” so=“” your=“” first=“” condition=“” is=“” met.=“” accordingly,=“” f’(x)=“” negative=“” because=“” f(x)=“” decreasing=“” g’(x)=“” 0=“” it=“” a=“” straight=“” line=“” so,=“”>any negative number
Therefore f(x)>g(x) while g’(x)>f’(x)</y<pi></p>
<p>oh, sorry… i misread the requirements -_-</p>
<p>Well, we can do Y=2^(-x) and Y=-1. 2^(-x) always > -1, but 2^-x is always decreasing so the derivative is always negative, and the deriv of a horiz. line is always 0.</p>
<p>Thanks for the -arc tan and y=-3. That one’s good.</p>
<p>My 2 cents.
Two functions whose range is all real numbers.
f(x) = x - e^(-x)
g(x) = x - 2e^(-x)
f(x) > g(x)
f’(x) < g’(x)</p>