For all real numbers x and y, let f be a function that f(x+y) =f(x)+f(y)+2xy and such that lim [ f(h)/ h ] = 7 (h-->0)
a) find f(0). Justify your answer
b) use the definition of the derivative to find f ' (x)
c) find f(x)
The graph of y = - 5/ (x-2) is concave down for all values of x such that
x<0 or x<2 or x< 5 or x>0 x> 2
Anonymous
Favorite Answer
a) The function takes two variables, but we can plug in some numbers to get f(0):
f(1) = f(1+0), so
f(1+0) = f(1) + f(0) +2(1)*(0), which means f(1) = f(1) + f(0), so f(0) = 0.
b) Remember the definition of the derivative is
Lim h->0 [ (f(x+h) - f(x)) / h ]
So using the definition of f(x+y), this is
Lim h->0 [ ( f(x) + f(h) + 2xh - f(x) ) / h ] =
Lim h->0 [ ( f(h) + 2xh ) / h ] =
Lim h->0 [ f(h)/h + 2x ] =
Lim h->0 [ f(h)/h ] + 2x =
7 + 2x
c) If we integrate this last expression with respect to x, we get
f(x) = x^2 + 7x + c. We know f(0)=0, so it's really
f(x) = x^2 + 7x
And just to check, notice that f(x+y) is
(x+y)^2 + 7(x+y) =
(x^2 + 2xy + y^2) + 7x + 7y =
(x^2 + 7x) + (y^2 + 7y) + 2xy = f(x) + f(y) + 2xy
Anonymous
the last one