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1. Where has the tangent to x3 - 5x got slope 4? Find the equations for the tangents at these points.
The derivative is
For slope 4, we need
So
So, the two x values where the slope is 4 are 31/2 and -31/2.
At 31/2, y = 31/2^3 - 5*31/2
Use your calculator to get a decimal answer, or use algebra to get y = 3*31/2 - 5*31/2 = -2*31/2.
Then substitute this into the equation for the tangent to get:
So
The equation of the tangent at 31/2 is:
| y = 4x - 6*31/2 |
At -31/2, y = 31/2^3 - 5*31/2
Use your calculator to get a decimal answer, or use algebra to get y = -3*31/2 + 5*31/2 = 2*31/2.
Then substitute this into the equation for the tangent to get:
So
The equation of the tangent at -31/2 is:
| y = 4x + 6*31/2 |
2.
Note, in the exam, I do not expect you to simplify the answers, just write them down.
a. (x3 + 9x2 + 7)/(3x2 - 9)
The derivative is
b. (x3 + x2 + x + 1)10 (2x3 + 3x2 + 5)11
The derivative is
c. ln(cos(ex))
The derivative is
d. e(x2 + x sin(x2 + 1))
The derivative is
e. (x3 + 3)(-1/3)
The derivative is
f. 8x + 7x
The derivative is
3. Sketch the graph of the derivative of the function sketched below.
A sketch of the derivative should look something like this:
4. Find the absolute maximum and minimum of the function f(x) = x5 - 2x3 + x + 1 on the interval [0,2]
We need to find the derivative first; this is
To find the critical values, we need to factor
So the critical numbers are 1/51/2, -1/51/2, 1, -1.
The interval is [0,2], so we don't need to worry about the negative critical numbers.
Make a table of the values of the function at the critical numbers and at the end points:
| 0 | 1 |
| 1/51/2 | 1.29 |
| 1 | 1 |
| 2 | 19 |
So the absolute maximum on the interval is y=19 at x=2, and the absolute minimum is y=1 at x=0, and also at x=1.
5. If the width of a square increases at 5m/s, and the length increases at 3m/s what is the rate of increase in the area when the width is 4m long, and the length 7m long?
Let the width be x, and the length be y, then the area is
Differentiate this relation, to get:
Then substitute in all the values, to get:
6. Find the intercepts, asymptotes, domain, local max and min, intervals where increasing, decreasing, concave up, concave down, of the function f(x) = (x3 + 1)/(x2 - 4). Use this information to sketch a graph of this function.
Domain This function has a real value unless x = 2 or -2, so the domain is (-infinity,-2),(-2,2),(2,infinity), in other words, all real numbers except 2 and -2.
y intercept: If x=0, y=1/(-4) = -1/4
The y intercept is at (0,-1/4)
x intercept:
If y = 0, 0 = (x3 + 1)/(x2 - 4), so (x3 + 1) = 0, so x3 = -1. The only solution to this is x = -1.
The x intercept is at (-1,0)
horizontal asymptote: If x is very large, y is approximately x^3/x^2 = x, which gets very large, so there is no horizontal asymptote.
vertical asymptote: to make y very large, we need the denominator close to zero, which means x is close to 2 or -2, so the vertical asymptotes are at
Local max and min: The derivative is:
To find where this is zero we need to factorise: f '(x) = [3x4 - 12x2 - 2x4 - 2x] /(x2 - 4)2
So one of the critical values is 0; the others could be found if you could factorize the cubic unfortunately this is not easy, so obviouly I've made a mistake in chosing this function to sketch the graph of. Ideally, it would factor, and then you could make a table and write down where the intervals of increase and decrease are.
There is not much I can do about this - I don't think there's much point in working through a complex example that I'm not expecting you to be able to do more of in terms of critical values and concavity; however, we can get some reasonable sketch by plotting a few more values: eg,
I appologise for this mistake - you can find better examples of this kind of question by looking in the text book, or looking at the links from handout 12, eg, hints4 page. Let me know if you need more examples.
Please send corrections to spelling mistakes, and any other comments, to me.
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