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 2x ² (x - 2)(x + 2)
Intercepts & Asymptotes

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 maximum at (0, 0) and local minimums 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 3(t - 1)(t - 3) = 0.
The particle is at rest at 1 second and at 3 seconds.

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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