max possible area / differentiable piecewise fcn

[xc630]

Hi I would appreciate some help with 3 problems.

1) a rectangle is to inscribed in a semicircle of radius 8 with one side lying on the diamter of the circle. What is the maximum possible area of the rectangle.

For this problem I do not know where to start so could someone get me started. How would I find where the rectangle and semicircle intersect?

2) For all x > 0, if f (ln x) = x^2, then f (x) =?

I took the E function of both sides and got x= e^ (x^2). This is not one of the options but am I along the right track?

3) If f (x) = { e^(-x)+2, for x <0 is differentiable at x=0 then a+b=?
ax+b, for x </= 0

The first function is 3 at x=0. What would I do next?

 

[galactus]

Re: max possible area

Hi I would appreciate some help with 3 problems.

1) a rectangle is to be inscribed in a semicircle of radius 8 with one side lying on the diameter of the circle. What is the maximum possible area of the rectangle.

For this problem I do not know where to start so could someone get me started. How would I find where the rectangle and semicircle intersect?

If the rectangle is centered at the origin, then its area is A=2xy.

The equation of the semicircle is \(\displaystyle \L\\y=\sqrt{64-x^{2}}\)

Therefore, \(\displaystyle \L\\A=2x\sqrt{64-x^{2}}\).

Find A'(x), set to 0 and solve for x.

 

[soroban]

Re: max possible area

Hello,xc630!

#2 is a strange problem!
I had to baby-talk my way through it.

2) For all \(\displaystyle x\,>\,0\), if \(\displaystyle f(\ln x) \:= \:x^2\), then \(\displaystyle f(x)\:=\:?\)

\(\displaystyle f(x)\) is a function such that, if we plug in \(\displaystyle \ln x\), we get \(\displaystyle x^2\)
What kind of function will do that?

. . . \(\displaystyle \ln x\;\;\Rightarrow\;\;\fbox{\begin{array}{c} \\ f(x) \\ \end{array}} \;\;\Rightarrow\;\;x^2\)

Here's the baby-talk . . .
\(\displaystyle f(x)\) must "square the \(\displaystyle x\)" ... then get rid of the log

We have: \(\displaystyle \:2\cdot\ln(x)\:=\:\ln(x^2)\)
. . So multiplying by 2 will "square the \(\displaystyle x\)".

To get rid of the log, exponentiate: \(\displaystyle \:e^{^{\ln(x^2)}} \:=\:x^2\)

Therefore: \(\displaystyle \:f(x)\:=\:e^{^{2x}}\)

 

[xc630]

thank you for the help!