Calculus 1300 Final Exam Review Problems

Solution:

First, notice that the expression is undefined at x=3 because the denominator becomes zero.

Factor the numerator: x - 9 = (x+3)(x-3)

The limit becomes: lim(x3) [(x+3)(x-3)/(x-3)]

Cancelling the (x-3) factors gives: lim(x3) (x+3)

Now we can directly substitute x=3: 3+3 = 6

Therefore, lim(x3) (x - 9)/(x - 3) = 6

Solution:

Rational functions are discontinuous where the denominator equals zero.

Set x - 1 = 0, which gives x = 1.

To verify this is a true discontinuity (not a removable one), check if (x - 4x + 3) = 0 at x = 1.

Evaluating: 1 - 4(1) + 3 = 1 - 4 + 3 = 0

Since both numerator and denominator equal zero at x = 1, we should check if the discontinuity is removable.

Factor the numerator: x - 4x + 3 = (x - 1)(x - 3)

The function can be written as f(x) = [(x - 1)(x - 3)]/(x - 1) = x - 3 for x 1

This is a removable discontinuity at x = 1. The limit exists but the function is undefined there.

Solution:

For the product rule: If f(x) = u(x)v(x), then f'(x) = u'(x)v(x) + u(x)v'(x)

Let u(x) = 3x - 2x, so u'(x) = 6x - 2

Let v(x) = sin(x), so v'(x) = cos(x)

Applying the product rule:

f'(x) = (6x - 2)sin(x) + (3x - 2x)cos(x)

This can be simplified if needed: f'(x) = sin(x)(6x - 2) + cos(x)(3x - 2x)

Solution:

For the chain rule: If f(x) = g(h(x)), then f'(x) = g'(h(x))h'(x)

Let g(u) = sin(u), so g'(u) = cos(u)

Let h(x) = 2x + 5x, so h'(x) = 6x + 5

Applying the chain rule:

f'(x) = cos(2x + 5x)(6x + 5)

f'(x) = (6x + 5)cos(2x + 5x)

Solution:

Let x be the width and y be the length of the rectangle.

Perimeter constraint: 2x + 2y = 120, so y = 60 - x

Area function: A = xy = x(60 - x) = 60x - x

To find the maximum, take the derivative and set it equal to zero:

A' = 60 - 2x = 0, which gives x = 30

Thus: y = 60 - 30 = 30

Therefore, the garden should be a square with dimensions 30 feet by 30 feet to maximize the area.

The maximum area is 30 30 = 900 square feet.

Solution:

Direct substitution gives 0/0, an indeterminate form, so we can apply L'Hpital's Rule.

Differentiate both numerator and denominator:

For the numerator sin(x), the derivative is cos(x)

For the denominator x, the derivative is 1

Applying L'Hpital's Rule:

lim(x0) (sin(x)/x) = lim(x0) (cos(x)/1) = cos(0)/1 = 1/1 = 1

Therefore, lim(x0) (sin(x)/x) = 1

Solution:

Using the power rule for integration: xdx = x^(n+1)/(n+1) + C

(3x + 2x - 5)dx = 3xdx + 2xdx - 51dx

= 3(x/3) + 2(x/2) - 5x + C

= x + x - 5x + C

Solution:

Let u = x, so du/dx = 2x, which means du = 2xdx or (1/2)du = xdx

Change of limits: when x = 0, u = 0 = 0; when x = 1, u = 1 = 1

Converting the integral:

[0,1] xe^(x)dx = (1/2)[0,1] e^u du

= (1/2)[e^u] = (1/2)(e - e) = (1/2)(e - 1)

Therefore, [0,1] xe^(x)dx = (e - 1)/2

Solution:

First, find the points of intersection by setting the equations equal:

x = 2x - x

2x - 2x = 0

2x(x - 1) = 0

So, x = 0 or x = 1

To determine which curve is on top, test a point between 0 and 1, like x = 0.5:

For y = x: y = (0.5) = 0.25

For y = 2x - x: y = 2(0.5) - (0.5) = 1 - 0.25 = 0.75

Therefore, y = 2x - x is on top between x = 0 and x = 1.

The area is: A = [0,1] [(2x - x) - x]dx

= [0,1] (2x - 2x)dx

= [x - (2/3)x]

= (1 - 2/3) - 0 = 1/3

The area bounded by the curves is 1/3 square units.

Solution:

When rotating about the x-axis, we can use the disk method.

The volume formula is: V = [a,b] [f(x)]dx

In this case, f(x) = x, with limits from x = 0 to x = 4.

V = [0,4] (x)dx

= [0,4] x dx

= [x/2]

= (4/2 - 0/2)

= (16/2)

= 8

Therefore, the volume of the solid is 8 cubic units.

Solution:

First, find the y-coordinate when x = 1:

y = 1 - 3(1) + 2 = 1 - 3 + 2 = 0

So the point is (1, 0).

Next, find the derivative of y:

y' = 3x - 3

Evaluate the derivative at x = 1:

y'(1) = 3(1) - 3 = 3 - 3 = 0

So the slope of the tangent line is 0.

The equation of the tangent line with slope 0 passing through (1, 0) is:

y - 0 = 0(x - 1)

y = 0

Therefore, the tangent line is the x-axis (y = 0).

Solution:

For limits at infinity of rational functions, the behavior is determined by the highest power terms.

Divide both numerator and denominator by x:

lim(x) (3x + 2x - 5)/(5x - x + 8)

= lim(x) (3 + 2/x - 5/x)/(5 - 1/x + 8/x)

As x , terms with x in the denominator approach 0:

= (3 + 0 - 0)/(5 - 0 + 0) = 3/5

Therefore, lim(x) (3x + 2x - 5)/(5x - x + 8) = 3/5

Solution:

Let u = ln(x), so du/dx = 1/x, which gives du = dx/x

The integral becomes:

(ln(x)/x)dx = u du = u/2 + C

Substituting back:

= [ln(x)]/2 + C