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Question: If the map $f: R \rightarrow R$ be defined by $f(x) = 4x - 1$ and the $g: R \rightarrow R$ be define...

If the map f:R→Rf: R \rightarrow R be defined by f(x)=4x−1f(x) = 4x - 1 and the g:R→Rg: R \rightarrow R be defined by g(x)=x3+2g(x) = x^3 + 2. What is value of (gof)x(gof)x?

A

64x3−48x2+12x+164x^3 - 48x^2 + 12x + 1

B

4x3−74x^3 - 7

C

64x3+48x2−12x+164x^3 + 48x^2 - 12x + 1

D

4x3+74x^3 + 7

Answer

64x3−48x2+12x+164x^3 - 48x^2 + 12x + 1

Explanation

Solution

The problem asks for the composition of two functions, (g∘f)(x)(g \circ f)(x), given f(x)f(x) and g(x)g(x).

The definition of the composition of functions (g∘f)(x)(g \circ f)(x) is g(f(x))g(f(x)).

Given functions are: f(x)=4x−1f(x) = 4x - 1 g(x)=x3+2g(x) = x^3 + 2

To find (g∘f)(x)(g \circ f)(x), we substitute f(x)f(x) into g(x)g(x): (g∘f)(x)=g(f(x))(g \circ f)(x) = g(f(x)) Substitute f(x)=4x−1f(x) = 4x - 1 into g(x)g(x): (g∘f)(x)=g(4x−1)(g \circ f)(x) = g(4x - 1)

Now, replace xx in the expression for g(x)g(x) with (4x−1)(4x - 1): g(x)=x3+2g(x) = x^3 + 2 So, g(4x−1)=(4x−1)3+2g(4x - 1) = (4x - 1)^3 + 2

Next, expand the term (4x−1)3(4x - 1)^3 using the binomial expansion formula (a−b)3=a3−3a2b+3ab2−b3(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3. Here, a=4xa = 4x and b=1b = 1. (4x−1)3=(4x)3−3(4x)2(1)+3(4x)(1)2−(1)3(4x - 1)^3 = (4x)^3 - 3(4x)^2(1) + 3(4x)(1)^2 - (1)^3 =64x3−3(16x2)(1)+3(4x)(1)−1= 64x^3 - 3(16x^2)(1) + 3(4x)(1) - 1 =64x3−48x2+12x−1= 64x^3 - 48x^2 + 12x - 1

Finally, substitute this back into the expression for (g∘f)(x)(g \circ f)(x): (g∘f)(x)=(64x3−48x2+12x−1)+2(g \circ f)(x) = (64x^3 - 48x^2 + 12x - 1) + 2 (g∘f)(x)=64x3−48x2+12x+1(g \circ f)(x) = 64x^3 - 48x^2 + 12x + 1