G
Guest
Guest
If f(x)= {(3,5), (2,4), (1,7)} and h(x)= {(3,2), (4,3), (1,6)} how in the world do you solve (f+h)(1)?
- Joined
- Feb 4, 2004
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- 16,582
You would solve in the usual way:
. . . . .(f + g)(x) = f(x) + g(x)
. . . . .(f - g)(x) = f(x) - g(x)
. . . . .(f × g)(x) = [f(x)]×[g(x)]
. . . . .(f ÷ g)(x) = [f(x)] ÷ [g(x)]
. . . . .(2f + 3g)(x) = 2×f(x) + 3×g(x)
...and so forth. So apply the definition of "f + h" to split the expression "(f + h)(1)" into to evaluations, plug "1" in for the x-value of each function, and read off the y-value of each function. Then add the two numbers.
Eliz.
. . . . .(f + g)(x) = f(x) + g(x)
. . . . .(f - g)(x) = f(x) - g(x)
. . . . .(f × g)(x) = [f(x)]×[g(x)]
. . . . .(f ÷ g)(x) = [f(x)] ÷ [g(x)]
. . . . .(2f + 3g)(x) = 2×f(x) + 3×g(x)
...and so forth. So apply the definition of "f + h" to split the expression "(f + h)(1)" into to evaluations, plug "1" in for the x-value of each function, and read off the y-value of each function. Then add the two numbers.
Eliz.
G
Guest
Guest
Thank YOu SO MUCH!!