Home > Precalculus > Function Composition

Let #h(x) = x/(x+4)# and #k(x)=2x-4#, how do you find (hºk)(x) and simplify?

Answer 1

Sophie Adams

#(h@k)(x) = (x-2)/(x)#

#(h@k)(x)# is the same as #h(k(x))#

In other words, simplify #h(2x-4)#.

To do this, we plug in #2x-4#, or replace every #x# in #h(x)# with #2x-4#.

#h(k(x)) = (2x-4)/((2x-4)+4)#

Then, we simplify:

#h(k(x)) = (2x-4)/((2x-4)+4)#

#h(k(x)) = (2x-4)/(2x-4+4)#

#h(k(x)) = (2x-4)/(2x)#

The numerator and denominator have a common factor of #2#, so we factor out #2# and cancel.

#h(k(x)) = (2x-4)/(2x)#

#h(k(x)) = (2(x-2))/(2(x))#

#h(k(x)) = (x-2)/(x)#

This is already simplified.

#(h@k)(x) = (x-2)/(x)#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Theodore Amidon

To find ( (h \circ k)(x) ), which represents the composition of functions ( h(x) ) and ( k(x) ), we substitute ( k(x) ) into ( h(x) ) wherever we see ( x ).

[ h(x) = \frac{x}{x+4} ] [ k(x) = 2x - 4 ]

Substituting ( k(x) ) into ( h(x) ):

[ h(k(x)) = \frac{2x - 4}{2x - 4 + 4} ]

Simplify:

[ h(k(x)) = \frac{2x - 4}{2x} ]

[ h(k(x)) = \frac{2(x - 2)}{2x} ]

[ h(k(x)) = \frac{x - 2}{x} ]

So, ( (h \circ k)(x) = \frac{x - 2}{x} ).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.