MAT 180: The SandBox

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91 thoughts on “MAT 180: The SandBox

    1. If \(f(0) = 1\) and \(f(2) = 3\) and \(f\) is exponential then \(f(x) = ab^x\) for some choice of constants \(a\) and \(b\). If \(f(0) = 1\) then \(ab^0 = 1.\) In which case, \(a = 1\). If \(f(2) = 3\) then \(b^2 = 3\) in which case \(b = \sqrt{3}\). Thus, \(f(x) = (\sqrt{3})^x = 3^{x/2}\).

  1. Professor Nevo, I’m not getting the correct answer for Problem 6. I combined the logs to get log(3-x)/(3+x)=2. Then I exponentiated both sides by 10 and simplified. From there on, I combined like terms and factored.

    1. If you exponentiate both sides, you get
      $$
      \frac{3-x}{3+x} = 10^2
      $$
      or
      $$
      3-x = 100(3+x)
      $$
      which is obtained by multiplying both sides by \(3+x.\)

      Now solve for \(x.\)

  2. Professor Nevo, I can’t seem to get e and f correct in Problem 2 on the WeBWorK. I’m not sure how to cancel out the natural logs to get simply x*y and x/y. Also how would I begin to solve Problem 3? Would exponentiating both sides help?

    1. For problem 3, take a look at Example 1.

      For problem 2 (e), note that if \(x = \ln A\) and \(y = \ln B\) then \(A = e^x\) and \(B = e^y\), by definition. In which case \(AB = e^x \cdot e^y\) or just \(e^{x+y}\). Use this substitution in 2 (f) as well and see what you get.

    1. Recall that the domain is the set of all values of \(x\) such that the function is defined. Logarithmic functions are not defined for values which make the inside part to logarithm negative. For example, \(f(x) = \log x\) is not defined for values of \(x\) less than or equal to zero. Therefore, the domain of \(f\) is \(x>0\) or in interval notation \((0, \infty)\). Another example: if \(g(x) = 2\log_2 (3x-1) +4\) then the domain is all values of \(x\) for which \(3x-1>0\) which is equivalent to \(x> 1/3,\) or in interval notation \((1/3, \infty).\)

    1. In general, to graph
      $$
      f(x) = a \cdot b^{kx-c} + d
      $$
      you can rely on three defining characteristics of the function:

      • The horizontal asymptote: \(y = d\)
      • The \(y\)-intercept (one always exists)
      • An additional point

      Plot the horizontal asymptote using a dashed line, plot the \(y\)-intercept and an additional point. Use a smooth curve to connect the two points and extend the graph so that the curve becomes asymptotic to the line \(y = d\). You should also extend the curve in the other direction and so that the curve moves away from the line \(y = d\). Essentially you want to preserve the general shape of the basic exponential function, when graphing its variations.

  3. Professor Nevo, Problem 6 is giving me trouble. I gave 180(1+.20)^n, as my answer but this was wrong. I’m not sure what else the answer could be.

    1. The area covered by a marsh, \(A,\) starts at 180 acres and drops by 1/5 each year for \(n\) years. So,

      Year Start Amount: End Amount:
      0 \(180\) \(\begin{aligned} & 180 – 180 \cdot 0.20 \\ = & 180(1-0.20) \end{aligned} \)
      1 \(180(1-0.20)\) \(180(1-0.20)^2\)
      2 \(180(1-0.20)^2\) \(180(1-0.20)^3\)
      3 \(180(1-0.20)^3\) \(180(1-0.20)^4\)
      $$\vdots$$ $$\vdots$$ $$\vdots$$
      n \(180(1-0.20)^n\) \(180(1-0.20)^{n+1}\)
  4. Problem 8 is giving me trouble again. I’m not sure how to rewrite 2cos^2t-1 in terms of cos(t)

    1. Use the double angle identity for cosine to obtain

      $$
      \cos4t = 2\cos^2(2t)-1
      $$

      You can’t stop here since the problem states that your solution must be in terms of \(\cos t \). So apply the double angle identity again:

      $$
      \begin{aligned}
      \cos4t & = 2\cos^2(2t)-1 \\
      & = 2(\cos 2t )^2 -1 \\
      & = 2(2\cos^2 t -1)^2 -1
      \end{aligned}
      $$

  5. Professor Nevo, for Problems 18-21 I’m not sure why I was wrong. For 18 my answer was (7/9)(sqrt(55)/8)+(-sqrt(32)/9)(-3/8) which was wrong, but I feel like I did everything I was supposed to do.

  6. Problem 4 Part 3 on the WeBWorK is giving me problems. The numerator should be sinxcosx+cosxsinx, however I’m not getting it correct.

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