Answer
(a) $f'(0) = 0$
$f'(\frac{1}{2}) = 1$
$f'(1) = 2$
$f'(2) = 4$
(b) $f'(-\frac{1}{2}) = -1$
$f'(-1) = -2$
$f'(-2) = -4$
(c) $f'(x) =2x$
(d) $f'(x) = 2x$
Work Step by Step
(a) $f(x) = x^2$
Using the graph, we can estimate the slope at the following points:
$f'(0) = 0$
$f'(\frac{1}{2}) = 1$
$f'(1) = 2$
$f'(2) = 4$
(b) We can see that the graph is even.
Then $f(-x) = f(x)$ and $f'(-x) = -f'(x)$
Using the graph and symmetry, we can estimate the slope at the following points:
$f'(-\frac{1}{2}) = -1$
$f'(-1) = -2$
$f'(-2) = -4$
(c) We could guess that $f'(x) = 2x$
(d) We can find an expression for $f'(x)$:
$f'(x) = \lim\limits_{h \to 0}\frac{(x+h)^2-x^2}{h}$
$f'(x) = \lim\limits_{h \to 0}\frac{(x^2+2xh+h^2)-x^2}{h}$
$f'(x) = \lim\limits_{h \to 0}\frac{2xh+h^2}{h}$
$f'(x) = \lim\limits_{h \to 0}(2x+h)$
$f'(x) = 2x$
