Como você encontra # (fg) (4) # dado #f (x) = 4x-3 # e #g (x) = x ^ 3 + 2x #?

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

#f(x)-g(x)=4x-3-x^3-2x#

#color(white)(f(x)-g(x))=2x-3-x^3#

#"to evaluate " (f-g)(4)" substitute x = 4 into " (f-g)#

#(f-g)(color(red)(4))=(2xxcolor(red)(4))-3-(color(red)(4))^3#

#color(white)((f-g)(4))=8-3-64#

#color(white)((f-g)(4))=-59#