Download Help Examples for w11 1. Given that f(x) = , then its domain is all

Help Examples for w11
1. Given that f (x) = (x+8)(x+4)(x−0)
, then its domain is all values of x except those
(x−6)2 (x+8)
that will make the denominator zero, that is, all real numbers except x = 6 and x = −8.
Since the factor (x + 8) cancels in this case, we get a hole at x = −8. The corresponding
y-coordinate of the hole is obtained by plugging in x = −8 in the function (x+4)(x−0)
(x−6)2
2
after the cancelation: y = (−8 + 4)(−8)/(−8 − 6) = 32/196. The vertical asymptotes
of the function come from (x+4)(x−0)
(don’t use (x + 8) since it has been canceled). So in
(x−6)2
this case it is only x = 6. The zeros (namely the x-intercepts) of the function come from
= 0 (again do not use (x + 8) in the numerator since x = −8 is
the equation (x+4)(x−0)
(x−6)2
excluded from the domain and the graph of the function does not have any point with
x = −8). So just solve (x + 4)(x − 0) = 0 (that is what you get once you multiply both
sides by the common denominator).
2. It is the same kind as 1.
3. Keep in mind that the non-vertical asymptote is the quotient of the rational
function (you need to perform a long division) if it turns out to be a linear equation.
4–9. Keep in mind of the steps to follow: (1) Set the inequality into equation and
solve the equation; (2) plot the solutions as circles or dots (depending on whether the
inequality has equal sign or not) on the number line in the increasing order; (3) test
each interval. If you do not know what I am talking about, you have been a really bad
student since I have talked about this many, many times in class and you cannot have
missed that if you paid attention. Read my old notes from last week.
10. Solve the following inequality. Write the answer in interval notation.
x−7
≥0
x + 18
Solution. As we talked about in class, this is solved almost like the ones from 4 to
9, with the exception that each denominator needs to be set to zero and the solution of
that need to be plotted on the number line as well.
x−7
(1) x+18
= 0. Multiply both sides by (x + 18): x − 7 = 0 so x = 7. This needs to
be plotted as a dot since the inequality has equal sign. Set x + 18 = 0. x = −18. This
needs to be plotted as a circle as it comes from the denominator. Plug in a number
from each interval (−∞, −18), (−18, 7] and [7, ∞) into the original inequality to test.
(−∞, −18) and [7, ∞) work out so they are the solutions.
11–12. Similar to 10.
13. Find the interval on the real number line for
√ which the radicand is nonnegative
(greater than or equal to zero) so that the radical 8x − 5 defines a real number.
Solution. Since you can only take square root of a number that is ≥ 0, you must
have 8x − 5 ≥ 0. But you should be able to solve this inequality by now.
14. Solve the inequality x3 −64x ≤ 0. Write the answer in interval notation. Note: If
the answer includes more than one interval write the intervals separated by the ”union”
symbol, U. If needed enter ∞ as infinity and −∞ as -infinity.
Solution. You solve this just like you did for 4–9 since there are no fractions. I hope
that you will not tell me that you cannot solve the equation x3 − 64x = 0? (Pull out
the factor x first if you have to get a hint from me to get started.)
15. It is similar to 10–13.