Solution: When y = 0, the equation becomes

The only positive solution is 1, so the intersection of C with the positive x-axis is the point (1, 0).

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Differentiating implicitly and using the Chain Rule, we obtain

Solving for we obtain

At the point (1, 0), we haveTherefore, the equation of the tangent line there is

Solution: (a) LetSince g is continuous, by the Intermediate Value Theorem there exists a number $c$ ∈ (−1, 0) such thatg is increasing, so $g(x)$ = 0 has a unique solution.

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(b) Differentiating using the quotient rule, we obtain

Therefore, the critical points are $x$ = 0 and $x$ =$c$,where $c$ is as in part (a). Since c < 0, observe that

On the other hand,so $f$(c) is the maximal value of $f$ on [−1, 0].

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Therefore,

It follows that on the interval [−1, 2], $f$ attains its absolute maximum (equal to 7/5)at 2 and its absolute minimum (equal to −1) at −1 and 0.

Solution: (a) We claim that $f(x)$ = $cx$ + $d$, for some constant $d$. To prove this, set $g(x)$ =$f(x)$ − $cx$. Then

for all $x$. By a corollary of the Mean Value Theorem, it follows that $g(x)$ = $d$, for some constant $d$, and all $x$. This proves our claim:

$f(x) = cx + d$

(b) Since by the same corollary as above for some constant $c$. But for all $x$. By part (a), it follows that $f(x) = x+d$,for some constant $d$. But

$−1 = f(0) = d,$

so

Solution: (a) Since

it follows that the only critical point is x = 1, and thaton (1,∞). Thus:

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• $f$ is increasing on (−∞, 1);

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• $f$ is decreasing on (1,∞);

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• $f$ has an absolute maximum at $x$ = 1 equal to $f$(1) = $e$.

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(b) Differentiating, we obtain

The solutions to the equation (henceonIt follows that:

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(c) Sinceit follows that f(x) → 0, as x → ∞. Thus y = 0 is a horizontal asymptote at both +∞ and −∞.

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(d) The graph:

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