Try our Free Online Math Solver!

Mathematics 106 Problem Set Solutions
1. Find an equation for the line with x intercept 20 and
yintercept 5. Sketch its graph.
Solution : It goes through (20, 0) and (0, 5), so it has slope
. Its equation
may be written as . Its graph may be drawn
by simply connecting the two intercepts with a straight line.
2. Factor x ^{3} + 3x^{2} − 34x − 120 completely.
Solution : Looking at the divisors of 120, we see −4 is a zero of the
polynomial, so
x−(−4) = x+4 is a factor. Factoring x +4 out (using long division , synthetic
division
or some other method ), we get x^{3} + 3x^{2} − 34x − 120 = (x + 4)(x^{2} − x − 30).
The second factor may be further factored at sight or by trial and error to get
x^{3}+3x^{2}−
34x − 120 = (x + 4)(x + 5)(x − 6).
3. For what values of x is negative?
Solution: We note the numerator is 0 when x = 0 and when x = −6 and the
denominator
is 0 when x = 2.
When x > 2, all the factors are positive so the quotient is positive.
When 0 < x < 2, x − 2 < 0, so (x − 2)^{3} < 0, but the other factors are positive
so the
quotient is negative.
When −6 < x < 0, x < 0, (x − 2)^{3} is still negative, while x + 6 > 0, so the
quotient is
positive .
When x < −6, x+6 < 0 and x and (x−2)^{3} remain negative, so the quotient is
negative.
We conclude is negative when x < −6 and when
0 < x < 2. We may describe
the set for which is negative as {xx < −6
or 0 < x < 2} = (−∞,−6) ∪ [ (0, 2).
4. Calculate .
Solution: .
5. The distance s (in feet) traversed along a straight
path by an object by time t (in seconds)
is given by the formula s = 8t^{3} + 5t^{2} + 2t.
(a) Find its average speed during the time interval 2 ≤ t ≤ 4.
Solution: Its average speed is feet per
second.
(b) Find its instantaneous speed when t = 3.
Solution: Its instantaneous speed is . s' = 24t^{2}+10t+2, so its
instantaneous
speed is 248 feet per second.
6. Let f(x) = x^{3} + 5x. Use the definition of a derivative to find f'(x).
Solution:
7. Calculate .
Solution:
8. Calculate
Solution:
9. Calculate .
Solution:
10. Write down a strategy for calculating derivatives .
Solution: See notes online.
Prev  Next 