BRIEF ANSWERS FOR TEST 1

• T or F?
  • F, the correct statement is [f^(-1)(x)]' = 1/f'(f^(-1)(x)). It is easy to contradict the statement on the test by taking f(x) = e^x, or f(x) = x^3.
  • F. The limit is zero. L'Hospital's rule does not apply because the limit is not in the form 0/0.
  • F. If x=0 and C=1, this equality asserts that ln(2)=1, which is clearly false.
  • T, since y=e^x is continuous.
• Evaluate ...
  • This limit is in the form 1^(infinity), which is indeterminate. L'Hospital's Rule applies to show that the limit is e^3.
  • If you draw the correct triangle it is easy to see that sec(tan^(-1)(x/2)) = square_root(x^2+4)/2.
• Integrate ...
  • Use the substitution u = tan(x). Then you get an answer of 2^(tan(x))/ln(2) + C.
  • Since 3-2t-t^2 = 2^2 - (t+1)^2, we can use the integration formula involving sin^(-1)(t) that we just learned. We get that (before evaluating at the bounds) the integral is 6(sin^(-1)((t+1)/2)). After evaluating we get just Pi.
• Solve the IVP. Put it into normal form. P(x) = 1/(xln(x)), so nu(x) = ln(x). Thus y(x) = (1/ln(x))[(1/2)ln^2(x) + C]. From y(e)=1 we calculate that C = 1/2. Thus y(x) = (1/2)[ln(x) + 1/ln(x)]. It follows that y(square_root(e)) = 5/4.
• The bounds of integration are -Pi/3 and +Pi/3. The integral of 8cos(x) - sec^2(x) on this interval is 6*square_root(3).