calculus; funtion and integration

Question

1.   Given f (x) = 3x-2 and g (x) = 3 / x, x ≠ 0,
                 Expressed in a similar form 〖(gf)〗 ^ (-1).
                 State the domain and range of functions 〖(gf)〗 ^ (-1).

 

2.    Integrade ∫ ▒ (4x ^ 3 - 3x +2) / ((2x-1) (x +2)) dx. using the method of partial fractions.

Answer 1

(1) f(x)=3x-2, g(x)=3/x can be interpreted in two ways:

(a) Let h(x)=fg(x)=(3x-2)(3/x)=9-6/x.

Let y=h(x)=9-6/x. So 6/x=9-y, x=6/(9-y).

Let j(y)=x=6/(9-y), so j(x)=h^-1(x)=6/(9-x).

Domain of j(x)=h^-1(x)=[fg(x)]^-1 is all x≠9, that is, {(-∞,9) (9,∞)}.

Range is (-∞,∞).

(b) fg(x) could mean f○g(x)=f(g(x))=f(3/x)=3(3/x)-2=9/x-2.

Let y=h(x)=9/x-2. So 9/x=y+2, x=9/(y+2).

Let j(y)=x=9/(y+2), so j(x)=h^-12x)=9/(x+2).

Domain of j(x)=h^-1(x)=[fg(x)]^-1 is all x≠-2, that is, {(-∞,-2) (-2,∞)}.

Range is (-∞,∞).

(2) 4x³-3x+2 has a higher degree than (2x-1)(x+2)=2x²+3x-2, so we need to divide first:

                  2x         -3 (quotient)

2x²+3x-2 ) 4x³        -3x+2

                  4x³+6x²-4x

                        -6x² +x +2

                        -6x² -9x+6

                               10x-4 (remainder)

CHECK: (2x-3)(2x²+3x-2)+10x-4=4x³+6x²-4x-6x²-9x+6+10x-4=4x³-3x+2✔️

Integrand becomes 2x-3+(10x-4)/((2x-1)(x+2)).

Now we can work out partial fractions:

(10x-4)/((2x-1)(x+2))≡A/(2x-1)+B/(x+2).

Ax+2A+2Bx-B≡10x-4⇒A+2B=10, 2A-B=-4, 4A-2B=-8, 5A=2, A=⅖, B=24/5.

(10x-4)/((2x-1)(x+2))≡⅖(1/(2x-1))+(24/5)(1/(x+2)).

Indefinite integration:

∫(2x-3+(10x-4)/((2x-1)(x+2)))dx=x²-3x+0.2ln|2x-1|+4.8ln|x+2|+C where C is the constant of integration.

If the integration is between limits, the question doesn't display what they are.

If the interval of integration is [a,b] then the result is:

(b-a)(a+b-3)+0.2ln((2b-1)/(2a-1))+4.8ln((b+2)/(a+2)).