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How would you complete the problems that involve f(x) and h(x). I have never seen this before.... what do I do with them?
Please do 1, 3, 5, and the first 7.
Thanks!!!!~!
How would you complete the problems that involve f(x) and h(x). I have never seen this before.... what do I do with them?
Please do 1, 3, 5, and the first 7.
Thanks!!!!~!
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1. (f+h)(1) = 6 + 7 = 13
3. f^-1(x) = {(5,3),(4,2),(7,1)} i.e. { (y,x) }
5. (f o g)(4) = f(g(4)) = f(sqrt(4-3)) = f(1) = 7
7. 1/f(x) = [f(x)]^-1 = { (3,1/5), (2,1/4), (1,1/7)}
4: k(x) = x^2 + 5 so x = sqrt(k - 5) => k^-1(x) = sqrt(x-5)
6: g(k) = sqrt(k-3) , k(7) = 7^2 +5 = 54
so g(k(7)) = sqrt(54 -3) = sqrt 51
2: (k-g)(5) = k(5) - g(5) = 30 - sqrt2
3. f^-1(x) = {(5,3),(4,2),(7,1)} i.e. { (y,x) }
5. (f o g)(4) = f(g(4)) = f(sqrt(4-3)) = f(1) = 7
7. 1/f(x) = [f(x)]^-1 = { (3,1/5), (2,1/4), (1,1/7)}
4: k(x) = x^2 + 5 so x = sqrt(k - 5) => k^-1(x) = sqrt(x-5)
6: g(k) = sqrt(k-3) , k(7) = 7^2 +5 = 54
so g(k(7)) = sqrt(54 -3) = sqrt 51
2: (k-g)(5) = k(5) - g(5) = 30 - sqrt2
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1.
(f + h)(1)
= f(1) + h(1) [find ordered pair with 1 as x-coordinate; f(1) is the corresponding y-coordinate]
= 7 + 6
= 13
3.
f^-1(x) [switch x and y coordinates]
= {(5, 3), (4, 2), (7, 1)}
5.
(f o g)(4)
= f(g(4)) [g(4) = sqrt(4 - 3) = sqrt(1) = 1]
= f(1)
= 7
7.
1/f(x) [reciprocate y coordinate]
= {(3, 1/5), (2, 1/4), (1, 1/7)}
(f + h)(1)
= f(1) + h(1) [find ordered pair with 1 as x-coordinate; f(1) is the corresponding y-coordinate]
= 7 + 6
= 13
3.
f^-1(x) [switch x and y coordinates]
= {(5, 3), (4, 2), (7, 1)}
5.
(f o g)(4)
= f(g(4)) [g(4) = sqrt(4 - 3) = sqrt(1) = 1]
= f(1)
= 7
7.
1/f(x) [reciprocate y coordinate]
= {(3, 1/5), (2, 1/4), (1, 1/7)}