Given:
f(x)=2x−5
g(x)=3x2−5x+2
h(x)=x3−x
a) In order to find g(5x+2) we must plug in for every value of x in g(x) the value (5x+2).
g(x)=3x2−5x+2
g(5x+2)=3(5x+2)2−5(5x+2)+2
Now we simplify the expression on the right hand side.
=3(5x+2)(5x+2)−5(5x+2)+2
We FOIL (5x+2)(5x+2) and distribute (−5) to (5x+2)
=3[(5x)(5x)+(5x)(+2)+(2)(5x)+(+2)(+2)]−25x−10+2
=3[25x2+10x+10x)+4]−25x−8
=75x2+60x+12−25x−8
g(5x+2)=75x2+35x+4
b) For this part, we again substitute the given value of 5 into h(x)
f(x)=2x−5
h(x)=x3−x
f(x)−h(5)=2x−5−[(5)3−(5)]
=2x−5−125+5
f(x)−h(5)=2x−125
c) For this part, we need only to perform arithmetic on the given functions:
f(x)=2x−5
g(x)=3x2−5x+2
h(x)=x3−x
f(x)[h(x)−g(x)]=(2x−5)[x3−x−(3x2−5x+2)]
Simplifying the expression in the brackets, we get
=(2x−5)[x3−x−3x2+5x−2]
=(2x−5)[x3−3x2+4x−2]
Now we need to expand and multiply (2x−5) by [x3−3x2+4x−2]
=(2x−5)[x3−3x2+4x−2]
We can distribute the terms in (2x−5) and multiply each of them by [x3−3x2+4x−2]
=(2x)[x3−3x2+4x−2]+(−5)[x3−3x2+4x−2]
This gives:
=2x4−6x3+8x2−4x
−5x3+15x2−20x+10
Combining like terms, gives:
=2x4+(−6−5)x3+(8+15)x2+(−4−20)x+10
f(x)[h(x)−g(x)]=2x4−11x3+23x2−24x+10
