1.4: Problem 2 (1 point) For the functions f(x) 4x2, g(x)-3-2, and h(x) = x2-4x

Question

1.4: Problem 2 (1 point) For the functions f(x) 4x2, g(x)-3-2, and h(x) = x2-4x + 1, find the following if they exist (f·g)(x) = (h . h)(x) = If any of the quantities do not exist, write "undefined" in the blank provided You have 2 attempt(s) remaining before you will receive a new version of this problem. Note: You can earn partial credit on this problem You have attempted this problem 15 times. Your overall recorded score is 67% You have unlimited attempts remaining. Page generated at 12/06/2017 at 12.24am CST WeBWork O 1996-2016I theme: math4 I ww.version: 2.121pg_version: 2.121 The WeBWorK Projedt

Explanation / Answer

f = 4x^2
g = 3x - 2
h = x^2 - 4x + 1

fog :
f[g(x)]
f(3x-2)
= 4(3x-2)^2 ---> ANS

gof :
g[f(x)]
= g(4x^2)
= 3(4x^2) - 2
= 12x^2 - 2 ---> ANS

hoh :
h[h(x)]
h(x^2 - 4x + 1)
= (x^2 - 4x + 1)^2 - 4(x^2 - 4x + 1) + 1

= x^4 + 16x^2 + 1 - 8x^3 - 8x + 2x^2 - 4x^2 + 16x - 4 + 1

= x^4 - 8x^3 + 14x^2 + 8x - 2 ---> ANS
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2)
Vertex = 4,5
So, we have

y = a(x - 4)^2 + 5

Now, plug in (7,6) to ifnd a :
6 = a(7-4)^2 + 5
a = 1/9

So, we have
y = 1/9*(x - 4)^2 + 5 ----> ANS

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3)
log(4/15)
= log4 - log15
= log(2^2) - log(15)
= 2log(2) - log(5*3)
= 2log(2) - log5 - log3
= 2x - q - S ---> ANS

log(30) :
log(2*3*5)
= log2 + log3 + log5
= x + q + S --> ANS

log(7) :
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4)
cos(alpha) = 8/9

cos(a/2) = +/- sqrt(1 + cos(a))/2

Now, a is in quad 4
So, a/2 is either in quad2 or quad3 where cos is negative

- sqrt(1 + (8/9)) / 2

= -sqrt(17/9) / 2

= -sqrt(17) / 6 ----> ANS

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