## Calculus: Early Transcendentals (2nd Edition)

$a.$ The value of the derivative is $f'(-1)=2$. $b.$ The equation is $y=2x+2$ $c.$ The graph is on the figure below.
$a.$ Using the definition of the derivative with $a=-1$ we have $$f'(-1)=\lim_{h\to0}\frac{f(-1+h)-f(-1)}{h}=\lim_{h\to0}\frac{1-(-1+h)^2-(1-(-1)^2)}{h}=\lim_{h\to0}\frac{1-(1+h^2-2h)-0}{h}=\lim_{h\to0}\frac{-h^2+2h}{h} = \lim_{h\to0}(-h+2)=-0+2=2.$$ $b.$ The equation of the tangent line through the point $(a,f(a))$ is given by $y-f(a)=f'(a)(x-a)$. Using $a=-1$, $f(a) =f(-1)= 1-(-1)^2=0$ and $f'(a)=f'(-1)=2$ we have $$y-0=2(x-(-1))\Rightarrow y=2x+2$$ $c.$ The graph is on the figure below. The function is graphed by the solid line and the tangent is dashed.