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Math 112 Solutions for 2.6, p. 167

1.
(a) about -1.97. (b) $ f '(1)=-2$. (c) $ f\;'(x)=\dfrac{-2}{x^{3}}$, so $ f\;'(1)=-2$. (d) $ y=3-2x$.

2.
(c) Assume that the function has value 0 at $ x=0$. The graph never straightens out, indicating nondifferentiability at the origin.

(d) Assume the function has value 0 at $ x=0$. The graph eventually flattens out, indicating that the derivative at 0 is 0.

3.
(c) $ \frac{1}{9}$. (e) $ f '(0)=1$.
4.
(b) $ f '(x)=2ax+b$. (c) $ -1/x^{2}$.
5.
(a) (i) The one-sided limits of $ f$ at 1 are both 1$ =f(1)$, so $ f$ is continuous at 1. (ii) $ f '(1)=\lim_{b\to 1}\dfrac{f(b)-f(1)}{b-1}$. The two one-sided limits of this expression are both 2, so $ f '(1)=2$.

(b) (i) The one-sided limits are not equal, so the function is not continuous at 1. Since it is not continuous at 1, it is not differentiable at 1.

(c) (i) The one-sided limits at -1 both have value 5, so $ f$ is continuous at 5. (ii) $ f\;'(-1)=\lim_{x\to -1}\dfrac{f(x)-f(-1)}{x+1}$. The two one-sided limits of this expression are both -2, so $ f$ is differentiable at -1 and the derivative is -2.

6.
Assume that $ f, g,$ and $ h$ all have value 0 at 0.

(a) $ \lim_{x\to 0} \sin\frac{1}{x}$ does not exist, so $ f$ is not continuous at 0.

(b) $ \lim_{x\to 0}
x\sin\frac{1}{x}=0=g(0)$, so $ g$ is continuous at 0. But $ g'(0)=\lim_{x\to
0}\frac{g(x)-g(0)}{x-0} =\lim_{x\to 0} \sin\frac{1}{x}$, which does not exist, so $ g$ is not differentiable at 0.

(c) $ \lim_{x\to 0}h(x)=0=h(0)$, so $ h$ is continuous at 0. $ h'(0)=\lim_{x\to
0}\frac{h(x)-h(0)}{x-0}=\lim_{x\to 0}x\sin\frac{1}{x}=0$, so $ h$ is differentiable at 0.

11.
$ \dfrac{dA}{dr}=2\pi r$
14.
$ \dfrac{dR}{dq}=460-.04q$ and $ \dfrac{dR}{dq}\Big\vert _{q=300}=448$.

17.
Derivative is slope, so the derivative of a linear function is the (constant) slope of its graph.

20.
(a) .33 (c) .12, .23, and .175




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Jason Grout 2003-02-07