Question #343475

Two functions f : R → R and g : R → R are defined by f(x) = 5x3 + 1 and g(x) = 2x − 3 for all x ∈ R.

Determine the inverse of (f -1 ◦ g) and (g ◦ f )(2) and ( f ◦ g)(2).


1
Expert's answer
2022-05-26T08:04:15-0400

Let f(x1)=f(x2).f(x_1)=f(x_2). It means that



5x13+1=5x23+15x_1^3+1=5x_2^3+1x13=x23x_1^3=x_2^3(x1x2)(x12+x1x2+x32)=0(x_1-x_2)(x_1^2+x_1x_2+x_3^2)=0x1x2=0x_1-x_2=0x1=x2x_1=x_2

The function f(x)=5x3+1f(x)=5x^3+1 is bijective (one-to-one ) from R\R to R.\R.



f(x)=5x3+1,xRf(x)=5x^3+1, x\in \Ry=5x3+1y=5x^3+1

Change xx and yy

x=5y3+1x=5y^3+1

Solve for yy

y3=x15y^3=\dfrac{x-1}{5}




y=x153y=\sqrt[3]{\dfrac{x-1}{5}}

Then


f1(x)=x153,xRf^{-1}(x)=\sqrt[3]{\dfrac{x-1}{5}}, x\in \R



a)


(f1g)=2x3153=2x453,xR(f^{-1}\circ g)=\sqrt[3]{\dfrac{2x-3-1}{5}}=\sqrt[3]{\dfrac{2x-4}{5}}, x\in \R


y=2x453y=\sqrt[3]{\dfrac{2x-4}{5}}

Change xx and yy

x=2y453x=\sqrt[3]{\dfrac{2y-4}{5}}

Solve for yy

2y4=5x32y-4=5x^3




y=52x3+2y=\dfrac{5}{2}x^3+2

Then


(f1g)1=52x3+2,xR(f^{-1}\circ g)^{-1}=\dfrac{5}{2}x^3+2, x\in \R



b)


(gf)(x)=2(5x3+1)3=10x31(g\circ f)(x)=2(5x^3+1)-3=10x^3-1




(gf)(2)=10(2)31=79(g\circ f)(2)=10(2)^3-1=79

c)


(fg)(x)=5(2x3)3+1(f\circ g)(x)=5(2x-3)^3+1

(fg)(2)=5(2(2)3)3+1=6(f\circ g)(2)=5(2(2)-3)^3+1=6


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