Assume that f is a continuous function from the real numbers to the real numbers such that for all numbers x,y, and a it is true that f(x+y)=f(x) + f(y) and that f(ax)=af(x). Further assume that f(1) = pi. Find f'(-1). Please explain your reasoning.
I am so confused. Ive read many times but can't figure out where or how to begin. Please help!
if f(ax) = af(x), f(x) = kx for some k.
check:
f(ak) = k(ax) = a(kx) = af(x)
f(x+y) = k(x+y) = kx+ky = f(x)+f(y)
So, since f(1) = pi, f(x) = pi*x
f'(-1) = pi
In fact, f'(x) = pi for any x.
To find f'(-1), we need to find the derivative of the function f at x = -1.
The given conditions suggest that f(x) is a linear function, specifically a linear function of the form f(x) = cx, where c is a constant.
Let's apply the given conditions to determine the value of c:
1. f(1) = pi
Plugging x = 1 into the function f(x) = cx, we have f(1) = c(1) = c.
Therefore, c = pi.
2. f(ax) = af(x)
Plugging x = 1 and a = -1 into the equation, we have f(-1) = -1 * f(1) = -1 * pi = -pi.
So, we have determined that the function f(x) is f(x) = pi * x.
Now, to find f'(-1), we need to find the derivative of f(x) = pi * x and evaluate it at x = -1.
The derivative of f(x) with respect to x is f'(x) = pi.
Therefore, f'(-1) = pi.
In conclusion, f'(-1) is equal to pi.