Please help, I'm in Pre-Calc, reviewing Algebra stuff.

The value of c is -1

g(x) = x² + 2x - 1, x ≥ -1

x must be greater than or equal -1 to make g(x) one-one function

Step-by-step explanation:

Let us use these notes to solve our question

∵ g(x) = x² + 2x +c, where x ≥ -1

∴ g(x) is quadratic function

∵ f(x) is the inverse of g(x)

∵ f(2) = 1

→ That means point (2, 1) lies on the graph of f(x)

∴ (2, 1) ∈ f(x)

→ Switch x and y to find the image of this point on g(x)

∴ (1, 2) lies on the graph of g(x)

∴ (1, 2) ∈ g(x)

→ T find c substitute x and g(x) by the coordinates of the point (1, 2)

∵ x = 1 and g(x) = 2

∴ 2 = (1)² + 2(1) + c

∴ 2 = 1 + 2 + c

∴ 2 = 3 + c

→ Subtract 3 from both sides to find c

∵ 2 - 3 = 3 - 3 + c

∴  -1 = c

∴ The value of c is -1

∴ g(x) = x² + 2x - 1

To find the x-coordinate of the vertex point of g(x) use this rule h = , where a is the coefficient of x² and b is the coefficient of x

∵ a = 1 and b = 2

∴ h =

∴ h = -1

→ By using the 3rd note above we must restrict the domain at the

  x-coordinate of its vertex

∵ The x-coordinate of g(x) is -1

∵ f(x) is the inverse of g(x)

∴ The domain of g(x) must be greater than or equal to -1

→ The domain is the values of x

∴ g(x) has an inverse if x ≥ -1

∴ x must be greater than or equal to -1 to make g(x) one-one function

step-by-step explanation:

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