Don't let composite functions scare you. The notation is trickier than the concept of plugging the result of one function into a second.
(f:h)(x) = f(h(x))
f(h(x)) = f(-x-1)
=-2(-x-1)-1
=2x+1
(f:h)(-3) = -5
Done stepwise,
h(-3) = -(-3)-1 = 2
plug 2 into f(x)
f(2) = -2(2)-1 = -5
2006-10-05 07:43:27
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answer #1
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answered by novangelis 7
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ok, two times the sum of two numbers is 4. we've 2 unknowns, we could call them "x" and "y" two times their sum = 4 so we are in a position to write 2(x+y) = 4 so all of us comprehend that x+y = 2 *** now all of us comprehend that the sum of their squares is fifty two so, x^2 + y^2 = fifty two @@@ nicely, from the *** equation, we are in a position to rearrange and discover that y = 2 - x so we could sub that equation into our @@@ equation we are going to get x^2 + (2-x)^2 = fifty two now strengthen x^2 + 4 - 4x + x^2 = fifty two convey mutually like words 2x^2 - 4x + 4 = fifty two convey each and every thing to a minimum of one ingredient, and use the quadratic formulation 2x^2 - 4x - 40 8 = 0 (4 +- sqrt(4 hundred) )/4 = a million +- 5 = 6, and -4 now, y = 2 - x so y could equivalent -4, or 6 so which you have 2 solutions for this: x = 6, y = -4, and x = -4, y = 6
2016-10-18 21:12:13
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answer #2
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answered by ? 4
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h(-3) = -(-3)-1 = 2, so f(h(-3)) = f(2) = -2(2)-1 = -5.
2006-10-05 07:35:57
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answer #3
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answered by James L 5
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56
2006-10-05 07:32:45
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answer #4
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answered by Anonymous
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when you are looking for (f o h) (-3), first plug the -3 into the h(x) problem and take the answer you get and plug it into the f(x) probem.
h(x) = -x -1
h(-3) = -(-3) -1
h(-3) = 3 - 1
h(-3) = 2
f(x) = -2x-1
(f o h) (-3) = -2(2)-1
(f o h) (-3) = -4 - 1
(f o h) (-3) = -5
2006-10-05 07:38:42
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answer #5
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answered by Anonymous
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You should try joining a math club at your college, that's what they're for to solve problems together
2006-10-05 07:38:30
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answer #6
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answered by babysweetvee 3
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Such a hard problem!!
You don't need that in RL!
2006-10-05 07:38:31
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answer #7
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answered by ~Peace~N~Love~ 3
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if the mind is confused consult your chi is what i do
2006-10-05 07:33:31
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answer #8
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answered by Anonymous
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