James Finch Math - College, section 1, Fall 2019 Instructor: Dr. Friendly
WebAssign Stewart - Calculus ET (Metric) 8/e (Homework) Current Score : 17 / 40
Due : Friday, November 8, 2019 21:46 EST
1. 1/1 points | Previous AnswersSCalcET8M 2.5.503.XP.MI.
Use continuity to evaluate the limit. 40 +
lim
x
40 + x
x Õ 9
437
Solution or Explanation Click to View Solution
2. 0/11 points | Previous AnswersSCalcET8M 2.1.AE.001.
EXAMPLE 1 Find an equation of the tangent line to the function y = 3x 2 at the point P(1, 3). SOLUTION We will be able to find an equation of the tangent line t as soon as we know its slope m. The difficulty is that we know only one point, P, on t, whereas we need two points to compute the slope. But observe that we can compute an approximation to m by choosing a nearby point Q(x, 3x 2) on the graph (as in the figure) and computing the slope mPQ of the secant line PQ. [A secant line, from the Latin word secans, meaning cutting, is a line that cuts (intersects) a curve more than once.] We choose x ≠ 1 so that Q ≠ P. Then, 2
mPQ = 3x − 3 . x − 1 For instance, for the point Q(1.5, 6.75) we have
Video Example
2
mPQ =
6.75 − 3
1.5 − 1
2
=
52
3.75
.5
= 2
7.5 .
The tables below show the values of mPQ for several values of x close to 1. The closer Q is to P, the closer x is to 1 and, it appears from the tables, the closer mPQ is to 2 tangent line t should be m = 2 x
m PQ
x
m PQ
2
9
0
3
1.5
7.5
.5
4.5
1.1
6.3
.9
5.700
1.01
6.030
.99
5.970
6 . This suggests that the slope of the
6 .
1.001 6.003 .999 5.997 We say that the slope of the tangent line is the limit of the slopes of the secant lines, and we express this symbolically by writing 2 lim mPQ = m and lim 3x − 3 = 2 Q Õ P x Õ 1 x − 1
6 .
Assuming that this is indeed the slope of the tangent line, we use the point-slope form of the equation of a line (see Appendix B) to write the equation of the tangent line through (1, 3) as y − 2
3 = 2
y = 2
6 x − 2
6 (x − 1) or
3 .
The graphs below illustrate the limiting process that occurs in this example. As Q approaches P along the graph, the corresponding secant lines rotate about P and approach the tangent line t.
3. 1/1 points | Previous AnswersSCalcET8M 2.2.VE.003.
Watch the video below then answer the question.
x − 8 is the line y = 7. x − 7
The vertical asymptote of the function True False
Solution or Explanation False
4. 0/4 points | Previous AnswersSCalcET8M 2.4.TEC.006.
Concept
In this module, you can explore the definition of a limit for several functions. The version of the definition that is most useful for this module is: Let f be a function defined on some open interval that contains a, except possibly at a itself. Then lim f(x) = L
x Õ a
0 < |x − a| < if for every number > 0 there is a number > 0 such that if
then |f(x) − L| < .
The definition of a limit can be thought of as a simple game. You claim that the limit is a number L. Someone gives you . You then have to choose a small (but non-zero) number such that for all x values within a distance of a, all f(x) values are within a distance of L. For limits at infinity, the definition we use is: Let f be a function defined on some interval (a, ∞). Then lim f(x) = L
x Õ ∞
means that for every > 0 there is a corresponding number N such that if x > N then |f(x) − L| < . Instructions
First, choose The Limit of a Function from the drop down menu in the upper left corner. Then choose a function from the drop down menu in the upper right corner. Now choose a value of a with the slider. This moves the black dot along the curve so that its x-coordinate matches the value you have chosen. Use the slider to choose some value of greater than zero. As you do, two parallel blue lines will appear, producing the interval (L − , L + ) where L is the value of the function at a (represented by the horizontal black line passing through the dot). Use the slider next to choose some value of greater than zero. You will see two vertical lines appear, one a distance to the right of the dot and the other a distance to the left of the dot. Now find a value of such that for x-values between the vertical lines, the y-values are between the blue lines (in other words, the curve should enter and exit the box formed by the lines through the sides rather than through the top or bottom). When you have found such a value of , the shading within the box will turn green (otherwise, it will be red). This is one possible value of for the precise definition of the limit for the given value of and the selected function. You can click the Zoom In button to get a better look and to set a more precise value of . To explore limits at infinity, choose Limits at Infinity from the drop down menu in the upper left corner. Then choose a function from the drop down menu in the upper right corner. In order to choose the limit of the function at infinity, slide the x max slider to the right to see what value the function eventually approaches. Then use the L slider to adjust the horizontal line. If the function simply levels off, align the horizontal line with the right end of the function's graph. If the function oscillates, try to move the horizontal line to the center of the oscillations at the right end of the function's graph. Use the slider to choose some value of greater than zero. As you do, two parallel blue lines will appear, producing the interval (L − , L + ) where L is the limit at infinity (represented by the horizontal line). Use the N slider next to choose some value of N. You will see a vertical line appear. Continue to slide the N slider to the right until you find a value of N such that all values to the right are between the blue lines. When you have found such a value of N, the shading within the box will turn green (otherwise, it will be red). This is one possible value of N for the precise definition of the limit for the given value of and the selected function. You can click the Zoom In button to get a better look and to set a more precise value for N. Simulation
Click here to access the TEC simulation. Exercise
Select the fourth function in "Limits at Infinity", y = sin(x)/(x + 1) + /12. (a) Change the graph view to 0 ≤ x ≤ 100 and use the vertical slider at the right to set an estimated limit value L. 4332
0.260
(b) What is the smallest value of N that guarantees that |f(x) − L| < 0.02 for all x > N? 32432
52.1
(c) Expand the graph view to 0 ≤ x ≤ 500 and zoom in. Revise your estimated limit value L, if necessary. Illustrate the definition of limit by finding values of N that correspond to = 0.008 and = 0.003. N 0.008
23432
124.7
0.003
33.3
350.9
Solution or Explanation (a) Use the vertical L-slider to obtain an estimate of L » 0.260. Remember to use the x max -slider to set x max = 100. (b) The smallest value of N is 52.1. This is obtained after setting = 0.02 and moving N from 0.0 to the right until the box becomes green for the first time, which is, in this case, N = 52.1. (c) After moving x max to 500 and then clicking Zoom In button, we see that we have to revise the estimate of L to obtain an better estimate L » 0.262. As in part (b), by first setting the value for and then moving N from 0.0 to the right until it hits the first green box, we can obtain the smallest values given as follows: N = 124.7 for = 0.008 and N = 350.9 for = 0.003.
5. 5/5 points | Previous AnswersSCalcET8M 2.5.502.XP.
Consider the following. ex if x < 1
f(x) =
x3
if x ≥ 1
, a = 1
Find the left-hand and right-hand limits at the given value of a.
e
lim f(x) = −
xÕ1
1
lim f(x) = +
xÕ1
Explain why the function is discontinuous at the given number a. Since these limits are not equal
not equal , lim f(x) does not exist
x Õ 1
does not exist and f is therefore discontinuous at 1.
Sketch the graph of the function.
Solution or Explanation f(x) =
ex if x < 1 x3
if x ≥ 1
The left-hand limit of f at a = 1 is lim f(x) = lim ex = e. The right-hand limit of f at a = 1 is lim f(x) = lim x 3 = 1. Since these limits are not equal, lim f(x) does not exist and f is x Õ 1−
x Õ 1−
x Õ 1+
x Õ 1+
x Õ 1
discontinuous at 1.
6. 2/2 points | Previous AnswersSCalcET8M QP.6.008.
(a) Sketch the line with slope
3 that passes through the point (−2, 1). 2
Flash Player version 10 or higher is required for this question. You can get Flash Player free from Adobe's website. Submission Data
line: y=3/2*x+4
(b) Find an equation for this line. y=32x+4
Read more about Topic 6: Linear Functions
7. –/2 pointsSCalcET8M A.A.020.
Solve the inequality in terms of intervals and illustrate the solution set on the real number line. 1 < 3x + 7 ≤ 28
Flash Player version 10 or higher is required for this question. You can get Flash Player free from Adobe's website.
8. 1/1 points | Previous AnswersSCalcET8M 5.5.502.XP.MI.
Evaluate the integral by making the given substitution. (Use C for the constant of integration.) x 3(1 + x 4)5 dx, u = 1 + x 4
124(1+x4)6+C
Solution or Explanation Click to View Solution
9. 0/1 points | Previous AnswersSCalcET8M A.E.020.
Write the sum in sigma notation. 5 − 5x + 5x 2 − 5x 3 + · · · + (−1)n5x n n j = 0
(−1)n5xn Solution or Explanation Click to View Solution
10. 1/2 points | Previous AnswersSCalcET8M 1.5.518.XP.
Find the exact value of each expression. (a) tan(arctan(8)) 8
(b) sin−1(sin(8/3)) 8.37758041
Solution or Explanation Click to View Solution
11. 1/1 points | Previous AnswersSCalcET8M 12.4.517.XP.
Find a unit vector orthogonal to both given vectors. i + j + k, 4i + k 1√26,3√26,−4√26
Solution or Explanation Click to View Solution
12. 0/1 points | Previous AnswersSCalcET8M 9.3.004.
Solve the differential equation. y' + 5xey = 0 y=ex
Solution or Explanation y' + 5xey = 0
dy
= −5xey
dx
e−y dy = −5x dx
e−y dy =
−5x dx
−e−y = −
5 2 x + C 2
5 e−y = x 2 − C 2
−y = ln
5 2 x − C 2
13. 0/2 points | Previous AnswersSCalcET8M 12.6.019.
Use traces to sketch the surface. y = 2z 2 − 2x 2
Identify the surface. hyperbolic paraboloid elliptic cylinder ellipsoid elliptic cone elliptic paraboloid hyperboloid of two sheets hyperboloid of one sheet parabolic cylinder
Solution or Explanation Click to View Solution
14. 5/6 points | Previous AnswersSCalcET8M 2.2.EI.001.
Review the Explore It, then use it to complete the exercise below. Explore It
CONCEPT
EXAMPLE VIDEO
EXPLORE & TEST
WHEN WOULD I USE THIS
YOU WILL LEARN ABOUT: How to find the limit of a function using its graph.
Click here to access the Explore It in a new window. Select Function 1, x ≤ −1
−0.5x − 2 2 − x 2 f(x) = 2
−1 < x < 1
2 − x 2
1 < x ≤ 2
x = 1
0.1(x − 2)3 − 2 x ≥ 2, under the Explore & Test section of the Explore It. (a) Does
lim f(x) exist?
x Õ −1−
Yes No
Compute the limit if it exists (round to four decimal places). (If the limit does not exist, enter DNE.) 1.5000
(b) Does
lim f(x) exist?
x Õ −1+
Yes No
Compute the limit if it exists (round to four decimal places). (If the limit does not exist, enter DNE.) 1.0000
(c) Does lim f(x) exist? x Õ −1
Yes No
Compute the limit if it exists (round to four decimal places). (If the limit does not exist, enter DNE.) DNE
y = −ln
5 2 x − C 2
