Let #f(x) = 4x-1, h(x) = x-2#. What is #(f*f)(0)#?

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Answer 1 · 2017-11-04T01:48:16

See a solution process below:

First, the function #h(x)# plays no role in this problem.

We can write #(f * f)(x)# as:

#(f * f)(x) = f(x) * f(x) = (4x - 1) * (4x - 1)#

Or

#(f * f)(x) = (4x - 1) * (4x - 1)#

To find #(f * f)(0)# we can substitute #color(red)(0)# for each occurrence of #color(red)(x)# in #(f * f)(x)# and calculate the result:

#(f * f)(color(red)(x)) = (4color(red)(x) - 1) * (4color(red)(x) - 1)# becomes:

#(f * f)(color(red)(x)) = ((4 * color(red)(0)) - 1) * ((4 * color(red)(0)) - 1)#

#(f * f)(color(red)(x)) = (0 - 1) * (0 - 1)#

#(f * f)(color(red)(x)) = -1 * -1#

#(f * f)(color(red)(x)) = 1#