Scribe Post: Day 65

Well it's Manny and I'm your scribe for today! Todays class we did on our pre-test on our fifth unit on applications of derivatives. Here's the questions with the answer, and what I think the solutions are to get the answer.

Multiple Choice Part

1) The Graph of the function f(x)= 2x^5/3 - 5x^2/3 is increasing on which of the following intervals?I. 1 < x II. 0 < x < 1 III. x < 0The answer is I. and III. only.
- We get this by differentiating f into f'(x) = (10/3)x^2/3 - (10/3)x ^-1/3.
- Then we can factor it into f'(x) = (10/3)x ^-1/3(x-1) by taking out the lowest common factor to find its roots in order to make a number line
- We now know that there is a root at 0, and 1, so where ever it is positive, the function f is increasing.

2) Let f(x) = x⁵ - 3x² + 4. For how many inputs c between a=-2 and b=2 is it true that f(b) - f(a) / b - a = f'(c)
The answer is 2.
- First we'd find f(b) and f(a), which equals 24, and -40 respectively.
- Then plug it into the equation to get f'(c) to equal 16.
- Then differentiate f to get f'(x) = 5x⁴ - 6x.
- And since we're looking at the point f'(c), and we know what the point is, we get 16 = 5x⁴ - 6x.
- Then we'd just solve for the roots and we get the answer for how many we would have.
- I'd just punch in the equation the calculator and look at it over the [-2,2] interval and look at the zeros.

3) The table below gives some values of the derivative of a function g.


Based on this information it appears that on the interval covered by the table...
The answer is that g has a point of inflection.
- From the information, we know that g will be increasing everywhere, since it g' is positive throughout the interval.
- But we notice it g' starts to decrease after increasing, so at the highest point before decreasing there is a horizontal tangent line which means that there is a point of inflection because there is a change of sign from a positive to negative rate of change.

4) Suppose f is a continuous and differentiable function on the interval [0,1] and g(x) = f(3x). The table below gives some values of f

What is the approximate value of g'(0.1)?
The answer is approx. 3.84.
- For this we use the chain rule law.
- We know that g(x) = f(3x).
- So then we find g'(x) = f'(3x)(3) by the chain rule.
- After just plug in the values to get g'((.1)) = f'(3(.1))(3).
- Which turns into g'((.1)) = f'(.3)(3), then solve using the symmetric difference quotient.

5) If f(x) = ln(x) - k√(x) has a local minimum at x = 4 then the value of k is:
The answer is 1.
- If x = 4 is a minimum on the graph of f, then f'(4) = 0 must be true.
- So differiantiate f into f'(x) = 1/x - (1/2)k(x)^-1/2
- Then plug in the 4 f'(4) = 1/4 - (1/2)k(4)^-1/2
- Now solve for k.


Long Answer Part

1) Let f be a function given by f(x)= (3x - 2) / ( √(2x² + 1) )

a) Find the domain of f.
The answer is (-∞,∞)
When deciding domain, we need look only at the denominator. And in this function the denominator will never equal 0, meaning it's never undefined which means the function won't have vertical asymptotes. From that we know that the domain includes all the real numbers.

b) On the graph below, sketch the graph of f.


- We can sketch this by first looking where the function has roots. Roots occur when the function equals 0, or undefined. Look at both the numerator and denominator for roots. In this function the denominator can never equal 0 because of the square root function, which means there is no vertical asymptote. But in the numerator we notice that there is a root at 2/3.

- Then we can look at where we have y-intercepts, which occur when x=0. By plugging the 0 in for x into our function, it is easily located.

- Next, we can look at if theres any horizontal asymptote. And there is since the highest degree of power in the numerator is equal to the degree in the denominator. Horizontal asymptotes are discovered by the limits of infinity concepts. From this information we can just take the leading coefficient in the numerator which is 3, and divide it to the leading coefficient in the denominator which is √(2), so there is one at 3/√(2). And we can't forget about the -∞ side, to find that there is another horizontal asymptote at -3/√(2).

- Then we need to find out where the function is increasing and decreasing, and its concavity. For this we need the derivative. But the derivative was already given in part (d). So from finding the root to be -3/4, we make a number line to see that it is, decreasing to the left of it, and increasing to the right, which means there's a minimum at that point.

c) Write an equation for each horizontal asymptote of the graph of f.
The answer is lim x->∞ f(x)= 3/√(2) and lim x->-∞ f(x)= -3/√(2).
- Applying the infinity concept to our function. The -2 and 1, in the numerator and denominator respectively are insignificant to infinity, so we disregard it.
- So the denominator can be rewritten as √(2) * √(x²), which is √(2)*(x).

d) Find the range of f. [ Use f'(x) = 4x + 3 / (2x² + 1)^3/2 to justify your answer]
The answer is [ -17/4√(17/8), 3√(2) ).
- Since there's a minimum at x = -3/4, we can just plug in that value in our function to get the minimum output.
- Our maximum output is approaching 3/√(2) but never actually touching because of the asymptote there, found in part c).

Click here (part 1, part 2)to look at the paper with the correct circle answers and other multiple choices.

And that was it for the whole day. *Whew!* I'm a bit still unsure about parts of the long answer part, sooo if you're confused, I am too =). I tried my best to answer these problems in full detail as much to my understanding. Our group did pretty poorly on the test afterall. I'm scared for the test *gulp..* Anyway.. next scribe for Thursday will be Danny. Later days!