1001 Algebra Problems.PDF

f(x) =

–(x– 1)^2 + 3 =

–(x^2 – 2x+ 1) + 3 =

• x^2 + 2x– 1 + 3 =


• x^2 – 2x+ 2

Finally, simplify the original expression

f(x+ h) – f(x):

f(x+ h) – f(x) =

(–x^2 – 2hx– h^2 + 2x+ 2h+ 2) – (–x^2 + 2x+ 2) =

• x^2 – 2hx– h^2 + 2x+ 2h + 2 + x^2 – 2x–2 =
  –2hx– h^2 + 2h=
  h(h– 2h+ 2)

696. c.By definition, (g ̊ h)(4) = g(h(4)). Observe
     that h(4) = 4 – 2 4 = 4 – 2(2) = 0, so g(h(4))
     = g(0) = 2(0)^2 – 0 – 1 = –1. Thus, we conclude
     that (g ̊ h)(4) = –1.


697. d.By definition, (f ̊ f ̊ f)(2x) = f(f(f( 2 x))).
     Working from the inside outward, we first
     note that f(2x) = –2(x)^2 = –4x^2. Then,f(f(2x))
     = f(–4x^2 ) = – (–4x^2 )^2 = –16x^4. Finally,f(f(f(2x)))
     = f(–16x^4 ) = – (–16x^4 )^2 = –256x^8. Thus, we
     conclude that (f ̊ f ̊ f)(2x) = –256x^8.


698. b.Begin with the innermost function: find
     f(–2) by substituting –2 for xin the function
     f(x):

f(–2) = 3(–2) + 2 = –6 + 2 = –4

Then, substitute the result for xin g(x).

g(–4) = 2(–4) – 3 = –8 – 3 = –11

Thus,g(f(–2)) = –11

699. e.Begin with the innermost function: Find

f(3) by substituting 3 for xin the function f(x):

f(3) = 2(3) + 1 = 6 + 1 = 7

Next, substitute that result for xin g(x).

g(7) = 7 – 2 = 5

Finally, substitute this result for xin f(x):

f(5) = 2(5) + 1 = 10 + 1 = 11

Thus,f(g(f(3))) = 11.

700. c.Begin with the innermost function. You are
     given the value off(x):f(x) = 6x+ 4. Substitute
     this expression for xin the equation g(x), and
     then simplify:

g(6x+ 4) =

(6x+ 4)^2 – 1 =

36 x^2 + 24x+ 24x+ 16 – 1 =

36 x^2 + 48x+ 15

Therefore,g(f(x)) = 36x^2 + 48x+ 15.

701. b.Since g(0) = 2 and f(2) = –1, we have

(f ̊ g)(0) = f(g)(0)) = f(2) = –1.

702. b.Since f(5) = 0 and f(0) = 0, we work from
     the inside outward to obtain

f(f(f(f(5)))) = f(f(f(0))) = f(f(0)) = f(0) = 0

703. c.Simplify the given expression:

f(x+ 2) =

(x+ 2)^2 – 4(x+ 2)=

x^2 + 4x+ 4 – 4 x– 8=

x^2 – 4

704. b.The domain ofg ̊ fconsists of only those
     values ofxfor which the quantity f(x) is
     defined (that is,xbelongs to the domain off)
     and for which f(x) belongs to the domain ofg.
     For the present scenario, the domain offcon-
     sists of only those x-values for which –3x 0,
     which is equivalent to x 0. Since the domain
     ofg(x) =  2 x^2 + 1 8 is the set of all real num-
     bers, it follows that all x-values in the interval
     (–∞, 0] are permissible inputs in the composi-
     tion function (g ̊ f)(x), and that, in fact, these
     are the only permissible inputs. Therefore, the
     domain ofg ̊ fis (–∞, 0].

ANSWERS & EXPLANATIONS–