Suppose f is a function which satisfies the properties:
f(x+y)=f(x)+f(y)+x2y+xy2
for all real numbers x and y, and
limx→0f(x)x=1.
f′(x)
thanks in advance!
To find the derivative of f(x), denoted as f'(x), using the given properties, we can apply the limit definition of a derivative.
The definition states that the derivative of a function f(x) at a point x is equal to the limit of the difference quotient as h approaches 0:
f'(x) = lim(h→0) [f(x+h) - f(x)] / h
Using the given property, f(x+y) = f(x) + f(y) + x^2y + xy^2, we can substitute x+h for y in the above equation:
f'(x) = lim(h→0) [f(x+(x+h)) - f(x)] / (x+h - x)
Expanding the equation:
f'(x) = lim(h→0) [f(2x+h) - f(x)] / h
Substituting the definition of f(x+y):
f'(x) = lim(h→0) [f(x) + f(2x) + x(2x + h) + x(2x + h)^2 - f(x)] / h
Canceling out the common term f(x):
f'(x) = lim(h→0) [f(2x) + x(2x + h) + x(2x + h)^2] / h
Expanding further:
f'(x) = lim(h→0) [f(2x) + 4x^2 + 4xh + h^2] / h
Taking out the common term:
f'(x) = lim(h→0) [f(2x) + 4x^2 + h(4x + h)] / h
As h approaches 0, the last term h(4x + h) approaches 0:
f'(x) = lim(h→0) [f(2x) + 4x^2] / h
Finally, taking the limit:
f'(x) = 4x^2 / 0
Since the denominator is approaching 0, this indicates that f'(x) is not defined for all x.
Therefore, using the given properties, we can conclude that the derivative f'(x) does not exist for the function f(x).