Consider the function defined as
f(x)=(x^2+2x+3)^2+4.
As x ranges over all real values, what is the minimum value of f(x)?
To find the minimum value of the function f(x), we need to determine the vertex of the corresponding parabola since the vertex represents the lowest point on the curve.
The given function is in the form f(x) = (x^2 + 2x + 3)^2 + 4, which is the square of another expression inside parentheses. To simplify the problem, let's refer to the expression inside the parentheses as g(x) = x^2 + 2x + 3.
To find the vertex of the parabola defined by g(x), we can use the formula x = -b/2a, where a and b are the coefficients of the quadratic term and the linear term, respectively, in the equation g(x) = ax^2 + bx + c.
In our case, the quadratic coefficient is a = 1 and the linear coefficient is b = 2. Plugging these values into the formula, we get x = -2/(2 × 1) = -1.
Now, substituting x = -1 back into the expression for g(x), we find g(-1) = (-1)^2 + 2(-1) + 3 = 1 - 2 + 3 = 2.
Therefore, the vertex of the parabola defined by g(x) is (-1, 2).
To determine the minimum value of f(x), we substitute x = -1 into the original function:
f(-1) = ((-1)^2 + 2(-1) + 3)^2 + 4
= (1 - 2 + 3)^2 + 4
= 2^2 + 4
= 4 + 4
= 8.
Hence, the minimum value of f(x) as x ranges over all real values is 8.