If the minimum value of f(x)=x^2+18x+a is the same as the minimum value of f(f(x)), what is the max value of a?
df/dx = 2x+18
dg/dx = (2f+18)(2x+18)
df/dx=0 when x = -9
dg/dx=0 when f = -9 or x = -9
f = -9 means
x^2+18x+a = -9
x = -9±√(72-a)
So, a <= 72
To find the maximum value of a, we need to determine the condition where the minimum value of f(x) is equal to the minimum value of f(f(x)).
Let's start by finding the minimum value of f(x). We can use the vertex form of a quadratic function, which is given by f(x) = a(x - h)^2 + k, where (h, k) represents the vertex of the parabola.
In this case, f(x) = x^2 + 18x + a. To find the vertex, we need to complete the square:
f(x) = (x^2 + 18x) + a
= (x^2 + 18x + 81) - 81 + a
= (x^2 + 18x + 81) - 81 + a
= (x + 9)^2 - 81 + a.
From this form, we can see that the vertex of the parabola is (-9, -81 + a). Therefore, the minimum value of f(x) occurs when x = -9, and the minimum value is -81 + a.
Next, let's consider f(f(x)), which represents substituting f(x) into the function f(x). To find the minimum value of f(f(x)), we need to substitute the value of the minimum of f(x) into the function f(f(x)), which gives us:
f(f(x)) = f(-9) = (-9 + 9)^2 - 81 + a = a - 81.
Since the minimum value of f(x) is equal to the minimum value of f(f(x)), we have -81 + a = a - 81.
By rearranging the equation, we find that a cancels out:
-81 + a - a = -81
-81 = -81.
This equation is true for any value of a. Therefore, there is no restriction or condition on a.
Hence, there is no maximum value of a. It can take any real value.