<p>If f’(x) and g’(x) exist, and f’(x)>g’(x) always, then f and g must intersect
A) Once
B) no more than once
The answer they say is B, and some people said 1/x vs 2/x, but the derivative doesnt exist at 0, so I was wondering everyones take on this.</p>
<p>I would say B.</p>
<p>My reasoning is say
f(x)=x^2+x
g(x)=x^2</p>
<p>f prime is 2x+1
g prime is 2x
therefore, they never intersect and fprime>gprime</p>
<p>I am not sure what the other answer choices are but I am guessing B in this case.</p>
<p>the graphs of f(x) and g(x) cant intersect, and neither can their derivatives.</p>
<p>f’(x)>g’(x)</p>
<p>therefore</p>
<p>f’(x) - g’(x) > 0</p>
<p>and</p>
<p>f’(x) - g’(x) = C(x), where C(x) is a guaranteed positive real for all x. Of course, the difference C depends on the value of x.</p>
<p>Assume C(x) = |G(x)|, the absolute value of G(x)</p>
<p>Integrating,</p>
<p>f(x) - g(x) = D(x) + Z, where D(x) is the antiderivative of C(x) and Z is the arbitrary constant.</p>
<p>Note: D(x) may be a piecewise function since there is no available antiderivative of an absolute value function.</p>
<p>If f(x) and g(x) intersect then f(x) - g(x) = 0</p>
<p>Therefore D(x) + Z = 0</p>
<p>D(x) = -Z</p>
<p>Now f(x) and g(x) are everywhere differentiable. If you know what this means, then, maintaining generality, you know that C(x) may not be everywhere differentiable because it is in essence an absolute value function. But that complicates things and so we’ll assume since this is BC that we’re dealing with “nice” functions.</p>
<p>If f(x) = e^(2x) + Z and g(x) = 0.5e^(2x) then C(x) = e^(2x) (obviously positive everywhere) and D(x) + Z = 0.5e^(2x) + Z.</p>
<p>If D(x) = Z then 0.5e^(2x) = -Z</p>
<p>obviously if z happens to be positive then the equality fails for our domain of real numbers. if z happens to be negative then the equality is satisfied.</p>
<p>hence the question is flawed because the equality depends especially on the range of the functions.</p>
<p>If you don’t trust me try it with different values of Z.</p>
<p>:)</p>