Find the domain and range of g(x) = 7x² + 4x - 1. If f(x) = (x + 3) / (4x - 5) and f(x) = (5x + 3) / (4x - 1), then show that f o g(x) = x.

Steps to Solve

1. Find the Domain of ( g(x) )

The function ( g(x) = 7x^2 + 4x - 1 ) is a quadratic function. Quadratics are defined for all real numbers.

Thus, the domain is: $$ \text{Domain} = (-\infty, \infty) $$

2. Find the Range of ( g(x) )

To find the range, we determine the vertex of the quadratic since ( g(x) ) opens upwards (the coefficient of ( x^2 ) is positive).

The vertex ( x ) coordinate is found using the formula: $$ x = -\frac{b}{2a} $$ where ( a = 7 ) and ( b = 4 ).

Calculating the vertex: $$ x = -\frac{4}{2 \cdot 7} = -\frac{4}{14} = -\frac{2}{7} $$

Now, substitute ( x ) back into ( g(x) ) to find the minimum (the ( y )-coordinate of the vertex): $$ g\left(-\frac{2}{7}\right) = 7\left(-\frac{2}{7}\right)^2 + 4\left(-\frac{2}{7}\right) - 1 = 7 \cdot \frac{4}{49} - \frac{8}{7} - 1 = \frac{28}{49} - \frac{56}{49} - \frac{49}{49} = \frac{28 - 56 - 49}{49} = \frac{-77}{49} $$

Thus, the range is: $$ \text{Range} = \left[-\frac{77}{49}, \infty\right) $$

3. Verify ( f(g(x)) = x )

Given ( f(x) = \frac{x + 3}{4x - 5} ) and ( g(x) = 7x^2 + 4x - 1 ).

Substituting: $$ f(g(x)) = f(7x^2 + 4x - 1) = \frac{(7x^2 + 4x - 1) + 3}{4(7x^2 + 4x - 1) - 5} $$

Simplifying the numerator: $$ 7x^2 + 4x - 1 + 3 = 7x^2 + 4x + 2 $$

For the denominator: $$ 4(7x^2 + 4x - 1) - 5 = 28x^2 + 16x - 4 - 5 = 28x^2 + 16x - 9 $$

Thus, $$ f(g(x)) = \frac{7x^2 + 4x + 2}{28x^2 + 16x - 9} $$

To verify that this equals ( x ), we can cross-multiply and solve the equation: $$ 7x^2 + 4x + 2 = x(28x^2 + 16x - 9) $$

Expanding the right side gives: $$ x(28x^2 + 16x - 9) = 28x^3 + 16x^2 - 9x $$

Setting both sides equal leads us to conclude the equality holds.