Since $f(x+h) = $ and $f(x) = $ , we can plug this results into the derivative definition: $\displaystyle f'(x) = \lim_{h \to 0} [f(x+h)-f(x)]/h$ $\displaystyle = \lim_{h \to 0} \biggl[\biggl($$\biggr) - \biggl($$\biggr)\biggr]/h$

Next, we simplify the numerator and then perform the division by $h$: $\displaystyle f'(x) =\lim_{h \to 0} \biggl[$ $\biggr]/h$ $\displaystyle =\lim_{h \to 0} $

We still have a limit in terms of $h$. However, the expression in the limit does does not depend on $h$. Therefore, the limit as $h$ goes to zero of this expression is different from the same as the expression itself. We can conclude that the derivative $f'(x)$ is $f'(x)=$ .

slope $= \biggl[\biggl($$\biggr) - \biggl($$\biggr)\biggr]\bigg/\biggl($$\biggr)$

Simplify your expression by multiplying out the quadratic: slope $= \biggl[$ $\biggr]\bigg/\biggl($$\biggr)$

Cancel the $h$ in the numerator and denominator to get your final answer: secant slope =

$\displaystyle \diff{f}{x} = \lim_{h \to 0 } \Bigl($ $\Bigr) = $

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Does the derivative $\diff{f}{x}\Bigr|_{x=-0.5}$ exist? yes no If so, what is its value? Is the function differentiable at $x=-0.5$? no yes

Does the derivative $\diff{f}{x}\Bigr|_{x=0}$ exist? no yes Is the function differentiable at $x=0$? yes no