Aug 8, 2024
f(x) = -7x + 5, evaluating f(2) involves replacing x with 2 to get -7*2 + 5 = -9.f(2) vs. y = -7x + 5 – Function notation is preferred for clarity and descriptiveness.*f(x) = -7x + 5 and g(x) = 3/2x - 1 are different functions despite using x as the variable.h(x) = -2x^2 + x - 1, evaluating h(-1) involves replacing x with -1.sqrt(x^2 - 3x) at x = 1 results in sqrt(1 - 3) = sqrt(-2), which is not a real number.g(x) = (2x + 1)/(x - 5), x = 5 is outside the domain because it makes the denominator zero.(f(x + h) - f(x)) / h.f(x + h) by replacing x with x + h.f(x + h) and f(x) in the difference quotient formula.h if possible.f(x) = x^2 + 1, f(x + h) = (x + h)^2 + 1 simplifies to x^2 + 2xh + h^2 + 1. The difference quotient then simplifies to 2x + h.5, 0, or -1 to understand the output.2x or x + 1 as inputs to test the function's flexibility.