MA 3972-MA-Book May 8, 2018 13:4682 STEP 4. Review the Knowledge You Need to Score High
Part B Calculators are permitted.- Enter into your calculatory 1 =| 2 x+ 4 |
andy 2 =10. Locate the intersection
points. They occur atx=−7 and 3. Note
thaty 1 is belowy 2 fromx=−7to3.
Since the inequality is less than or equal
to, the solution is− 7 ≤x≤3. (See
Figure 5.8-3.)
[−10, 10] by [−10, 15]
Figure 5.8-3- Enter in your calculatory 1 =x^3 − 2 xand
y 2 =1. Find the intersection points. The
points are located atx=−1,− 0 .618, and
1.618. Sincey 1 is abovey 2 in the
intervals− 1 <x<− 0 .618 and
x> 1 .618 excluding the endpoints, the
solutions to the inequality are the
intervals− 1 <x<− 0 .618 and
x> 1 .618. (See Figure 5.8-4.)
[−2, 2] by [−2, 2]
Figure 5.8-4- Enter tan
(
arccos√
2
2)
into your
calculator and obtain 1. (Note that
arccos√
2
2=
π
4
and tan(
π
4)
=1.)- Factore^2 x− 6 ex+ 5 =0as
(ex−5)(ex−1)=0. Thus, (ex−5)=0or
(ex−1)=0, resulting inex=5 and
ex=1. Taking the natural log of both
sides yields ln(ex)=ln 5≈ 1 .609 and
ln(ex)=ln 1=0. Therefore to the nearest
thousandth,x= 1 .609 or 0. (Note that
ln(ex)=x.)- The equation 3 ln 2x− 3 =12 is
equivalent to ln 2x=5. Therefore,
eln 2x=e^5 ,2x=e^5 ≈ 148 .413159, and
x≈ 74 .207. - Entery 1 =
2 x− 1
x+ 1
andy 2 =1 into your
calculator. Note thaty 1 is belowy 2 = 1
on the interval (−1, 2). Since the
inequality is≤, which includes the
endpoint atx=2, the solution is (−1, 2].
(See Figure 5.8-5.)
[−4, 4] by [−4, 7]
Figure 5.8-5- Examinef(−x) and f(−x)=−2(−x)^4
+(−x)^2 + 5 =− 2 x^4 +x^2 + 5 = f(x).
Therefore, f(x) is an even function.
(Note that the graph of f(x)is
symmetrical with respect to they-axis;
thus, f(x) is an even function.) (See
Figure 5.8-6.)
[−4, 4] by [−4, 7]
Figure 5.8-6- Entery 1 =x^4 − 4 x^3 into your calculator
and examine the graph. Note that the
graph is decreasing on the interval
(−∞, 3) and increasing on (3,∞). The
function crosses thex-axis at 0 and 4.
Thus, the zeros of the function are 0 and