Question

Could someone please help??? is the graph of f(x)=2(x^1/2)-2A, where A>0, increasing, decreasing, convex, concave, have an inflection pt at x=2A^(-2/3), 4A^(-2/3) or A^(-2/3)? I'm thinking that it's concave, but I was wondering if someone could check it just ot make sure. THanks!

Answer 1

increasing, solve
\[\frac{1}{\sqrt{x}}-A>0\] for \(x\)

Answer 2

oh that was wrong!!!

Answer 3

\[\frac{1}{\sqrt{x}}>A\]
\[\sqrt{x}<\frac{1}{A}\]
\[x<\frac{1}{A^2}\] that is better
of course also \(x>0\) but we knew that to begin with

Answer 4

second derivative has no A in it
it is just as before
\[f''(x)=-\frac{1}{x^{\frac{3}{2}}}\] and this is always negative

Answer 5

so always concave down (i guess that is convex)

Answer 6

oh i see you changed the problem, so maybe my first answer is incorrect
i wasn't reading carefully

Answer 7

huh? I'm really confused, sorry.

Answer 8

but i think you made a typo, because
\[f(x)=2\sqrt{x}-2A\] has exactly the same behaviour as \(f(x)=2\sqrt{x}\)
the \(-2A\) makes no difference

Answer 9

Oh i did, sorry! I meant -Ax

Answer 10

So i typed this into an online calculator and it as concave down, but i that decreasing because it not the first derivative? or is it concave down? I get confused with that stuff

Answer 11

ok so we take the first derivative, which we already did

Answer 12

that is for increasing and decreasing
that is what i solved above

Answer 13

then the second derivative has no A in it, the second derivative is just
\[f''(x)=-\frac{1}{x^{\frac{3}{2}}}\]

Answer 14

the A is gone, so the concavity is independent of A
the second derivative is negative for all values of \(x\)
therefore the function is always concave down

Answer 15

ok, thanks! that makes sense

Answer 16

I have one more question. Find f'(1) if f(x)=the integral (1+(X^3))^(1/2). Is this a trick question? cause it's asking for the derivative, and the integral is already in derivative form, would i just plug in 1 and get the square root of 2?