MATH 161 TEST 3 REVIEW

1.

(x-1)(x+3)2 (x-1)(x2 +6x +9) x3 +6x2 +9x - x2 – 6x – 9 x3 +5x2 +3x – 9

2.

X4 – 9x2>0

x2(x2-9)>0

x2(x+3)(x-3)>0 (-,-3)(3,)

3.

(-7,0),(8,0)

(-7,0) because it has an odd exponent

4. (x+3)(x-3)

X = -3,3

y = 0

none because you can’t solve the top for x

1/9 plug 0 in for x

(-7,0),(8,0)

(-7,0) because it has an odd exponent

(8,0) because it has an even

exponent

-3, 3

y = 0

1/9 plug zero in for x

5.

𝑥 =3𝑦

𝑦+5 𝑥(𝑦 + 5) = 3𝑦

𝑥𝑦 + 5𝑥 = 3𝑦 𝑥𝑦 − 3𝑦 = −5𝑥 𝑦(𝑥 − 3) = −5𝑥 𝑓−1 =−5𝑥

𝑥−3

6.

-3,1

Even exponent

Even exponent

7.

(x+3)(x-3)(x-8)

-3 1

(0+3)2 (0-1)2 = 9

X4

3 (one less than degree)

8.

0

y 2

x -4

0

0

2

-4

9.

-1 1 4

-2 0 3 5

- + - +

Less than – use the negatives (-,-1)(1,4)

10.

Shifts right one and up 1 ( - negative in front flips the graph)

11.

(x+3)(x-3)

since y intercept is -7/9 the graph is below x axis

-3, 3

0

12.

Slide and divide x2 – 41x – 42

(x – 42)(x+1) then divide x = 7, -1/6

x≠7, -1/6

13.

Shifts up 3 and right 1

14.

Since the exponents of x are the same

15.

Crosses at all zeros

positive Xs

(c) 3x(x-1)(x-2) is x(x)(x) = +x

(e) 3x(1-x)(2-x) is x(-x)(-x) = +x

(f) -x(1-x)(x-2) is –x(-x)(x) = +x

x -1/6, 7

-19

10

16.

Because squaring under square root is positive #

**Square 2 then add 2 to get 6

*ALWAYS SQUARE THE # THEN ADD IT TO ITSELF

EX :√√𝑥 − 7 −7 72 +7 = 56 THEN Domain {x|x 56}

17.

(x-2)(x-3)2

(√𝑥 − 2)2 + 5 = x+3

√𝑥2 + 5 − 2 = √𝑥2 + 3

(𝑥2 + 5)2 + 5 = x4+10x2+30

√√𝑥 − 2 − 2

x 2 (from original)

x6 (square the # then add it to the answer)

18.

Change the x and y values.

(0,-4) Label a point

(-4,0) on original.

Find the inverse (-4,0)

on the red function

19.

Since it follows negative x3 graph

we start from top left

-x3

--3,2

18

--3,2

2

2

-3 1

touches

crosses

20.

Domain and range switch

21. If f is in quadrant II, what quadrant is f -1 in?

Put a point in quadrant II then switch x and y and see what

quadrant the inverse coordinate is in.

(-2,1)

(1,-2) Quadrant IV

*Quadrants I and III inverse coordinates do not move!

22. Find the inverse of f(x) = 3x

x+5

Switch x and y then solve for y. 𝑥 =3𝑦

𝑦+5 𝑐𝑟𝑜𝑠𝑠 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 (𝑦 + 5) = 3𝑦

𝑥𝑦 + 5𝑥 = 3𝑦 𝑥𝑦 − 3𝑦 = −5𝑥 𝑦(𝑥 − 3) = −5𝑥

𝑓−1 =−5𝑥

𝑥 − 3

Domain of f is {x|x≠-5} Range of f is {y|y≠3}

Domain and Range of function and its inverse are opposite, THEREFORE…

Domain of f-1 is {x|x≠3} Range of f-1 is {y|y≠-5}

[-3, )

[1, )

23.

Switch x and y then solve for y. 𝑥 =−4𝑦+9

𝑦+4 𝑐𝑟𝑜𝑠𝑠 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 x(y+4) = -4y+9

𝑥𝑦 + 4𝑥 = −4𝑦 + 9 𝑥𝑦 + 4𝑦 = −4𝑥 + 9 𝑦(𝑥 + 4) = −4𝑥 + 9

𝑓−1 =−4𝑥 + 9

𝑥 + 4

Domain of f is {x|x≠-4} Range of f is {y|y≠4}

Domain and Range of function and its inverse are opposite, THEREFORE…

Domain of f-1 is {x|x≠4} Range of f-1 is {y|y≠-4}

24.

Because there is a break in the graph

25. Solve the inequality x4>16x2 solve first then factor

x4-16x2 > 0

x2(x2-16)>0

x2(x+4)(x-4)>0

*easiest to look at the graph of x4 and touches at 0 crosses at -4 and 4

We are looking for above the x-axis since it is > 0

(-,-4)(4, )

26. 𝑓(𝑥) =2−𝑥5

3

Polynomial degree: 5

Standard form: −𝑥5

3+

2

3

Leading term: −𝑥5

3 with constant:

2

3

27. Solve the inequality (𝑥−9)(𝑥+2)

(𝑥−2)< 0

*we check all the critical points on a number line x = -2, 0, 9

all are open because <

-2 0 2 9

check #s -3 0 3 10

- + - + we want all negative intervals

(-,-3)(2,9)