CALCULUS I XMAS REVIEW SOLUTIONS |

__QUESTION 1__

a) zeros: *x = -2*, *x = 1* and *x = 2*

b) , the equation of the tangent is *y = -4x + 4*

c) Since lines must be parallel, slopes must be equal,

so we set *f '(x) = slope = 1 *

d)

The curve is increasing on the intervals .

The curve is decreasing on the interval

e) The endpoints of the interval are at **(0, 4)** and **(3, 10)**.

From part (d), we know that the critical point at is on the interval [0, 3].

Since *f(1.54) = -1.59*, we have an **Absolute Min of -1.59** and an **Absolute Max of 10.**

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__QUESTION 2__

Given: *g(x) = 2x ^{ 4} - 8x^{ 2}* which factors as

Zeros are at *x = -2, x = 0 *and* x = 2*.

There are no Asymptotes -- it's a polynomial function.

__First Derivative: Maximum, Minimum, Increasing, Decreasing__

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This means we have critical points at

**Increasing **:

**Decreasing** :

Therefore, we have a **local max**imum at (0, 0) and **local min**imums at

__Second Derivative: Concave Down, Concave Up__

This means we have **points of inflection** at

**Concave Up** : *x < -0.82 *and* x > 0.82*

**Concave Down** : *-0.82 < x < 0.82*.

**Symmetry **: to the *y-axis*.

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__QUESTION 3__

Since *V = x² h = 4*, we know that *h = 4 / x²*.

Now, the surface area of the box *A = x² + 4xh* and when we substitute *h = 4 / x²* we get

The solution is to make the square base *2 dm* and the height *1 dm*.

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__QUESTION 4__

* y = x² + 2x*

a)

b) P(-3, 3) so *y' = -4*, the slope of the tangent is *-4*.

c) Since *y = 24 *at* x = 4*, and *y' = 10 *at* x = 4*, the tangent equation is *y = 10x - 16*.

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__QUESTION 5__

We have

a) __domain__ : *x > 3*;

__range__ : *y > 0*

b)

Now we rationalize the numerator, collect terms, and limit *h* to 0:

c) At *x = 7, y = ½, *and *y' = -1 / 16*

The equation of the tangent to this curve at (7, ½) is .

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__QUESTION 6__

a) | b) | c) |

d) | e) |

f)

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__QUESTION 7__

*s(t) = t ^{ 3} - 6t^{ 2} + 9t*

a) *v(t) = s'(t) = 3t ^{ 2} - 12t + 9*

b) *v(2) = - 3 metres / sec*

c) *s'(t) = v(t) = 0 *means that* 3t ^{ 2} - 12t + 9 = 0* so

The particle is at rest at

d) The particle is moving in a positive direction when *v(t) > 0*

This is when *0 < t < 1 *and* t > 3*.

e) s(0) = 0 | s(1) = 5 | s(3) = 0 | s(5) = 74 |

f) The particle travelled from 0 to 5 then back to 0 then out to 74

for a total of 84 metres in the first 5 seconds. (first 2 moves cover 10 metres)

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