Answer
a) f'(0)= 0
f'($\frac{1}{2}$)= .5
f'(1)= 1
f'(2)= 2
b)f'($-\frac{1}{2}$)= -.5
f'(-1)= -1
f'(-2)= -2
c) See graph
d) f'(x)=2x
Work Step by Step
a) In a graphing calculator, type the equation f(x)=$\frac{1}{2}$$x^{2}$ into the y= menu. Graph the equation. In the calculate menu, select the derivative option, or "dy/dx" and insert the x value.
b) Since the graph of the equation is a parabola with its vertex on x=0, inputs with the same absolute values have the same derivative with opposite signs e.g. f'(1)=1, f'(-1)=-1. In this specific equation, the input is also the derivative.
c) As the original function is a parabola, that means that the derivative must be linear. Since the derivative of the equation is 2x, the line must have a slope of 2.
d) The definition of a derivative is as follows:
f'(x)=$\lim\limits_{h \to 0}$ $\frac{f(x+h)-f(x)}{h}$
Now plug in the original equation, $x^{2}$
$\lim\limits_{h \to 0}$ $\frac{(x+h)^{2}-f(x)^{2}}{h}$
$\lim\limits_{h \to 0}$ $\frac{(x^{2}+2xh+h^{2})-x^{2}}{h}$
$\lim\limits_{h \to 0}$ $\frac{2xh+h^{2}}{h}$
$\lim\limits_{h \to 0}$ $\frac{h(2x+h)}{h}$
$\lim\limits_{h \to 0}$ (2x+h)
2x
