How do you find #(f-g)(3)# given #f(x)=x^2-1 and #g(x)=2x-3# and #h(x)=1-4x#?

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Answer 1 · 2017-04-03T16:08:41

Compute #(f-g)(x)# and then evaluate at #x = 3#

OR

Evaluate #f(3)" and "g(3)# and then perform the subtraction.

#(f-g)(3) = f(3)-g(3)#

Compute #(f-g)(x)#:

#(f-g)(x) = f(x) - g(x)#

Substitute the equivalents for #f(x)" and " g(x)#

#(f-g)(x) = x^2-1 - (2x-3)#

Distribute the minus sign:

#(f-g)(x) = x^2-1 -2x+3#

Combine like terms:

#(f-g)(x) = x^2 -2x+2#

Evaluate at #x = 3#:

#(f-g)(3) = 3^2 -2(3)+2#

#(f-g)(3) = 5#

OR

Evaluate #f(x)" at "x=3#:

#f(3) = #3^2 - 1#

#f(3) = 8#

Evaluate #g(x)" at "x=3#

#g(3) = 2(3)-3#

#g(3) = 3#

#(f-g)(3) = f(3)-g(3)#

#(f-g)(3) = 8-3#

#(f-g)(3) = 5#

Either always works.