<p>Let f be a continuous function with domain x>0 and let F be the function given by F(x) = integral from 1 to x of f(t)dt for x>0. Supposte that F(ab) = F(a) + F(b) for all a>0 and b>0 and that F'(1) = 3. </p>
<p>a) find f(1)</p>
<p>3???</p>
<p>b) prove that aF'(ax)=F'(x) for every positive constant a.</p>
<p>WHAT???? need help</p>
<p>c) Use the results from parts a and b to find f(x). Justify your answer.</p>
<p>?.............</p>
<p>bump bump bump</p>
<p>part a is simple:</p>
<p>F'(x) = d/dx[the integral from 1 to x of f(t)dt]</p>
<p>which, by the second part of the fundamental theorem of calculus is just:</p>
<p>f(x)</p>
<p>thus</p>
<p>f(1) = 3 = F'(1)</p>
<p>b) i can't quite get this one because i always end up with aF'(x)</p>
<p>are you sure that it is aF'(xa) and not just F'(xa)?</p>
<p>i think i have a and b
i jst don't know how to explain c</p>
<p>"F(a*b)=F(a)+F(B)"
Doesn't this look like a logarithmic identity?</p>
<p>Here is a more formal approach:
f(1)=3
aF'(ax)=F'(x) ==> a<em>f(a</em>x)=f(x)
Let x=1, a*f(a)=f(1)=3
f(a)=3/a ==> f(t)=3/t</p>