Assignment 28

course Mth 271

028. `query 28*********************************************

Question: `qQuery 3.7.12 sketch y = -x^3+3x^2+9x-2 **** Describe your graph, including a description of any intercepts, extrema, asymptotes and concavity.

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Your solution:

Find zeros

X= -2, therefore x+2 is a factor

Find other factor

Get equation

X^2 -5x +1

Therefore:

x^3 + 2 x^2 + 9 x - 2 = - (x + 2)*(x^2 – 5*x + 1).

This expression is zero when x+2 = 0 or when x^2 - 5 x + 1 = 0.

The first equation gives us x = -2.

To solve the second, use the quadratic formula.

x = [ -(-5) +- `sqrt{ (-5)^2 - 4 * 1 * 1 } ] / (2 * 1) = [ 5 +- `sqrt{21} ] / 2.

Simplify

x = .21 and x= 4.7 appr.

Use first and second derivative to find extrema.

Derv is -3x^2+6x + 9

Set = 0 to find critical numbers

Solve for x.

-3x^2+6x + 9 = 0

x^2 - 2x - 3 = 0

(x-3)(x+1) = 0

x = 3 or x = -1.

2nd derv is -6x + 6

Put 3 and -1 in for x and solve

Its neg when x is 3 therefore max where x=3 (3,25)

Its post when -1 therefore min where x = - 1 (-1,-7)

Set 2nd derv = 0

-6x + 6 =0

X=1 …. Here, the derv changes from post to neg, therefore there is a point of inflection at x=1 (1, 9)

The derivative is post therefore the function is concave up on (-infinity, 1).

The derivative is neg therefore the function is concave down on (1, infinity).

There are no vertical or horizontal asymptotes.

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Given Solution:

`a First we find the zeros:

You can find the zero at x = -2 by inspection (i.e., try a few simple values of x and see if you 'hit' one).

Knowing that x = -2 is a zero, you know that (x - (-2) ) = x + 2 is a factor of the expression. You can find the other factor by long division of x + 2 into -x^3 + 3 x^2 + 9x - 2. You get quotient x^2 - 5 x + 1.

Thus -x^3 + 2 x^2 + 9 x - 2 = - (x + 2)•(x^2 - 5•x + 1).

This expression is zero when x+2 = 0 or when x^2 - 5 x + 1 = 0.

The first equation we already know gives us x = -2.

The second is solved by the quadratic formula. We get

x = [ -(-5) +- `sqrt( (-5)^2 - 4 * 1 * 1 ) ] / (2 * 1) = [ 5 +- `sqrt(21) ] / 2.

Simplifying we get approximate x values .21 and 4.7.

Then we find maxima and minima using 1st and 2d derivative:

The derivative is -3x^2+6x + 9, which gives the equation -3x^2+6x + 9 = 0 for critical points. Dividing thru by -3 we get

x^2 - 2x - 3 = 0 or(x-3)(x+1) = 0 so

x = 3 or x = -1.

Second derivative is -6x + 6, which is negative when x = 3 and positive when x = -1.

At x = 3 we have a maximum. Evaluating y = -x^3+3x^2+9x-2 at x = 3 we get y = 25. The negative second derivative indicates that (3,25) is a maximum.

At x = -1 we have a minimum. Evaluating y = -x^3+3x^2+9x-2 at x = -1 we get y = -7. The positive second derivative indicates that this (-1,-7) is a minimum.

Finally we analyze 2d derivative for concavity and pts of inflection:

The second derivative -6x + 6 is zero when x = 1; at this point the derivative, which is linear in x, changes from positive to negative. Thus the x = 1 point (1, 9) is a point of inflection.

The derivative is positive and the function therefore concave up on (-infinity, 1).

The derivative is negative and the function therefore concave down on (1, infinity).

The function is defined for all x so there are no vertical asymptotes.

As | x | -> infinity the magnitude of the function -> infinity so there are no horizontal asymptotes. **

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Question: `qQuery 3.7.32 sketch (x^2+1)/(x^2-1)

Note: The problem in the text might be (x^2+1)/(x^2-2). If so the solution given below can be easily adapted to that function.

Describe your graph, including a description of any intercepts, extrema, asymptotes and concavity.

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Your solution:

Find zeros and intercepts

Function has no zeros (doesn’t cross x axis)

When x=0

y = (0^2+1)/(0^2-1) = -1

so the y intercept is (0,-1).

Find max and mins

Derv= - 4x/(x^2 - 1)^2,

Zero: x = 0.

So (0, -1) is the only critical point.

2nd derv is 4(3x^2 + 1)/(x^2 - 1)^3,

which is negative when x = 0. So the critical point gives us a max.

look at 2nd derv for concavity:

2nd derv =4(3x^2 + 1)/(x^2 - 1)^3

numerator is always positive,

denominator is negative where x^2 - 1 < 0, which occurs between x = -1 and x = 1.

So the 2d derivative is positive on (-infinity, -1) and on (1, infinity), where the graph will be concave up, and negative on (-1, 1), where the graph will be concave down.

Check for asymptotes:

The denominator of (x^2+1)/(x^2-1) is zero and the numerator isn't when x = +1 and also when x = -1. So there are vertical asymptotes at x = +1 and at x = -1.

}For very large positive x or for very large negative x the +1 in the numerator and the -1 in the denominator are both insignificant and the function value is very close to x^2 / x^2 = 1. The function approaches its horizontal asymptote y = 1 for both large positive x and large negative x.

Find where the function is post and neg

For x < -1 the function cannot change sign, since it is continuous and has no zeros. Testing any x < -1 gave me a positive value. On this interval the function is therefore positive.

Also true for x > 1.

For -1 < x < 1 the same argument shows that the function is negative

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Given Solution:

`a First we look for zeros and intercepts:

The numerator is never zero, being the sum of the positive number 1 and the nonnegative quantity x^2. So the function has no zeros, i.e., never crosses the x axis.

When x=0 we have y = (0^2+1)/(0^2-1) = -1 so the y intercept is (0,-1).

Next we analyze the derivative to see if we can find relative maxima and minima:

The derivative is - 4x/(x^2 - 1)^2, which has its only zero when x = 0. So (0, -1) is the only critical point.

The second derivative is 4(3x^2 + 1)/(x^2 - 1)^3, which is negative when x = 0. So the critical point gives us a maximum.

We analyze the second derivative to determine concavity:

The second derivative 4(3x^2 + 1)/(x^2 - 1)^3 has a numerator which is always positive, since x^2 is always positive. The denominator is negative where x^2 - 1 < 0, which occurs between x = -1 and x = 1.

So the 2d derivative is positive on (-infinity, -1) and on (1, infinity), where the graph will be concave up, and negative on (-1, 1), where the graph will be concave down.

Now we look for vertical and horizontal asymptotes:

The denominator of (x^2+1)/(x^2-1) is zero and the numerator isn't when x = +1 and also when x = -1. So we have vertical asymptotes at x = +1 and at x = -1.

}For very large positive x or for very large negative x the +1 in the numerator and the -1 in the denominator are both insignificant and the function value is very close to x^2 / x^2 = 1. The function approaches its horizontal asymptote y = 1 for both large positive x and large negative x.

We finally determine where the function is positive and where negative:

For x < -1 the function cannot change sign, since it is continuous and has no zeros. Testing any x < -1 gives us a positive value. On this interval the function is therefore positive.

The same is true for x > 1.

For -1 < x < 1 the same argument shows that the function is negative**

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