If #f(2x+1)=2x-1# for all real numbers x, then #f(x)# = ?

i have no idea how to do this, please help!

1 Answer
LM
Apr 29, 2018

#f(x) = x - 2#

Explanation:

the difference between #2x+1# and #2x-1# is #2#.

#(2x-1) - (2x + 1) = 2x-2x-1-1 = -2#

#(2x-1) - (2x + 1) = -2#

when #2x+1# is the input of the equation, #2x - 1# is the output.

this means that when #x# is the input, then #x - 2# is the output.

this can also be seen on a graph:

desmos.com/calculator

the graph in red is the graph of #f(x)#, where #f(x) = x - 2#

the graph in blue is the graph of #f(2x+1)#, where #f(x) = x-2#.

if you look at the blue graph along the #x#-axis from #1# to #2#, then you'll see that the graph goes across by #1# unit, #(1 rarr 2)# and up by #2# units #(1 rarr 3)#.

the gradient of the blue graph is therefore #2/1#, which is #2#.

it can also be seen that the blue line crosses the #y#-axis when #y# is at #-1#. this means that the #y#-intercept of the graph is #-1#.

all linear graphs can be written in #y = mx + c# form, where #m# is the gradient of the graph and #c# is the #y#-intercept.

here, #m = 2# and #c = -1#, so the equation of the graph in blue is #y = 2x-1#.

the blue line is the graph of #f(2x+1)#, so here, the equation of #f(2x + 1) = 2x - 1#.

this is only true when #f(x) = x - 2#, which is the graph shown in red.

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