The appropriate function is f(x) = (x−1)2 −9 I won’t sketch the graph, but I
describe it in sufficient detail to enable you to graph it. The line -solver.html">graph is a
that opens up and has its vertex at (1, -9). Its y- intercept is f (0) = 1 − 9 =
To find its x-intercepts, solve the equation
(x − 1)2 = 9
x − 1 = ±3
so x = −2 or x = 4.
Put your solutions in interval notation here:
(1) (x − 1)2 − 9 < 0
(2) (x − 1)2 − 9 ≥0
9. Let f(x) = 3x − 1 and g(x) = x2 + 1. Evaluate the
ex pression ()(x).
()(x) = g(f(x)) = g(3x − 1) = (3x − 1)2 + 1 = 9x2 − 6x + 1 + 1.
10. Let f(x) = |x|, let g(x) = x−7, and let h(x) = x2. Write each of the
functions as a composition of functions chosen from f, g, and h.
(a) P(x) = |x − 14| − 7
This function subtracts x from its argument twice (), takes the
absolute value () and then subtracts 7 again ()
(b) N(x) = (|x| − 7)2
This function takes the absolute value of its argument (f), substracts 7
() and then squares ().
11. Find the inverses of the following functions:
(a) f(x) = 2x + 5
The inverse function will reverse these actions as follows
(1) subtract 5 from its input
(2) divide the result by 2
Clearly the inverse function is the function
and use the switch and solve method. After switching, solve for y as
12. Show that the function given by
To do this we need to show that
implies that a = b.
Simplify the equation as follows:
13. Show that the function given by y = |x + 2| is not one-to-one.
We need to show that there are two different values x = a and x = b such that
|a + 2| = |b + 2|.
We can think of such values by contemplating the meaning of an absolute value.
Now |x+2| stands for the distance of x from -2. We need to find different
that are the same distance away from -2. To do this, imagine -2 on the number
line, and take one step to the right arriving at -1, or one step to the left,
at -3. Note that
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