Biology Forums - Study Force

1. Find the domain of  f(x)=1/sqrt(3+2x)
2. If f(x)=1-x^2 and g(x)=1/sqrt x, find (f*g)(x)
3. An open box is to be made from a rectangular piece of material 9 in.  12 in. by cutting equal squares from each corner and turning up the sides. Let x be the length of each side of the square cut out of each corner. Write the volume V of the box as a function of x.
4. solve for x: 2^(3x-1)=5
5. solve for x: (x-2)^(3/2)=8
6. Use a reference triangle to find the exact value of cos[arctan(-3/10)]
7. Given y=arctan(1/x), find cos y.
8. solve for x: ln(5x-1)-lnx=3

\(\frac{1}{\sqrt{3+2x}}\)

\(\sqrt{3+2x}\ \ne 0\)

Solve for x:

\(x\ne -\frac{3}{2}\)

Also, inside the radical, it needs to be greater or equal to 0:

\(3+2x\ge 0\)

\(x\ge -\frac{3}{2}\)

\(\therefore D=\left\{x\mid x>-\frac{3}{2},\ x∈\Re \right\}\)

We're finding f∘g, so insert the content of g into f:

\(f\left(\frac{1}{\sqrt{x}}\right)=1-\left(\frac{1}{\sqrt{x}}\right)^2\)

Now simplify:

\(f\left(\frac{1}{\sqrt{x}}\right)=1-\frac{1}{x}\)

Simplify more if needed:

\(f\left(\frac{1}{\sqrt{x}}\right)=\frac{x-1}{x}\)

Recall that volume is length * width * height.

\(v=l\times w\times h\)

\(\therefore \ f\left(x\right)=v=\left(12-2x\right)\left(9-2x\right)x\)

Expand the factors:

\(\therefore \ f\left(x\right)=\left(12-2x\right)\left(9x-2x^2\right)=4x^3-42x^2+108x\)

\(2^{3x-1}=5\)

Use natural log or log base whatever, easier to use ln:

\(\ln \left(2^{3x-1}\right)=\ln \left(5\right)\)

\(\left(3x-1\right)\ln \left(2\right)=\ln \left(5\right)\)

Divide both sides by ln:

\(x=\frac{\frac{\ln \left(5\right)}{\ln \left(2\right)}+1}{3}\ =1.107\)

The reported answer is an approximate, you may have to round it based on your teacher's recommendations.

\(\left(x-2\right)^{\frac{3}{2}}=8\)

Take \(\ln\) of both sides:

\(\left(\frac{3}{2}\right)\ln \left(x-2\right)=\ln 8\)

Divide both sides by 3/2:

\(\ln \left(x-2\right)=\frac{\ln 8}{\frac{3}{2}}\)

\(\ln \left(x-2\right)=\frac{2\cdot \ln 8}{3}\)

Raise both sides as exponents to the base e:

\(e^{\ln \left(x-2\right)}=e^{\frac{2\cdot \ln 8}{3}}\)

\(x-2=e^{\frac{2\cdot \ln 8}{3}}\)

Bring the 2 over, and you're done:

\(x=e^{\frac{2\cdot \ln 8}{3}}+2=6\)

Answer's 6.

\(\cos \left(y\right)=\frac{x}{\sqrt{x^2+1}}\)

Combine using rules of logs:

\(\ln \left(\frac{5x-1}{x}\right)=3\)

Raise both as exponents to the base e, this gets rid of ln:

\(e^{\ln \left(\frac{5x-1}{x}\right)}=e^3\)

\(\frac{5x-1}{x}=e^3\)

\(5-\frac{1}{x}=e^3\)

Bring 5 and divide both sides by -1:

\(\frac{1}{x}=5-e^3\)

\(\frac{1}{5-e^3}=x=-0.06628\)

Now check your answer, plug back into original, and it should give you 3!