Let h(x)=4x2+2x+5.
Let g(x)=x^(1/4) +3
What is g(x + 1)?
Thank you
2007-02-05 15:17:50 · 4 answers · asked by J R 2 in Science & Mathematics ➔ Mathematics
Note how in the second formula x is in the parentheses...You want to know what g(x+1) is...
So if g(x)=x^(1/4)+3 then g(x+1)=(x+1)^(1/4)+3
h(x) is unneeded information
2007-02-05 15:21:37 · answer #1 · answered by Michelle M 2 · 0⤊ 1⤋
1. Solve g(x) = x^(1/4) + 3 with g(x + 1)
First: replace "x+1" with the x-variable in the function...
g(x) = x^(1/4) + 3
g(x+1) = (x+1)^(1/4) + 3
Sec: raise both terms in parenthesis to the power of "1/4"...
g(x+1) = x^(1/4) + 1^(1/4) + 3
g(x+1) = x^(1/4) + 1/4 + 3
Third: combine the last terms...
g(x+1) = x^(1/4) + 13/4
2007-02-06 00:29:08 · answer #2 · answered by ♪♥Annie♥♪ 6 · 0⤊ 0⤋
x^(5/4) + 3
2007-02-05 23:20:26 · answer #3 · answered by Kipper to the CUP! 6 · 0⤊ 1⤋
g(x)=x^(1/4) + 3
g(x+1)=(x+1)^(1/4) + 3
2007-02-05 23:22:40 · answer #4 · answered by seah 7 · 0⤊ 1⤋