What Calculus Concepts Are Tested in These Sample Problems?

  • Thread starter Gauss177
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But yes, you are correct in saying that the first derivative can also be used to classify local extrema.
  • #1
Gauss177
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Sorry for so many questions, I'm just trying to understand everything before my test coming up soon.

1. Let R be the region in the first quadrant under the graph of y=1/sqrt(x) for
4<=x<=9.

a) If the line x=k divides the region R into two regions of equal area, what is the value of k?
My thoughts: For this one, I know how to find the area of R, but I don't know how to start finding the line that divides the region into multiple parts of equal area.

2. Let f be a function given by f(x)=x^3 - 6x^2 + p, where p is an arbitrary constant.

a) Write an expression for f '(x) and use it to find the relative maximum and minimum values of f in terms of p. Show the analysis that leads to your conclusion.
My thoughts: f '(x) = 3x^2 - 12x. Setting this equal to 0 and solving for x gives: x=0, 4. These are critical points, and I know that to find max/min points I need to look at second derivative. But here it says just use first derivative? I'm also not sure what to write for analysis part.

b) For what values of the constant p does f have 3 distinct real roots?
I don't have any thoughts on this one :frown:

c) Find the value of p such that the average value of f over the closed interval [-1, 2] is 1.
 
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  • #2
Gauss177 said:
1. Let R be the region in the first quadrant under the graph of y=1/sqrt(x) for
4<=x<=9.
From the question you know that;

[tex]\int_{4}^{k}\frac{dx}{\sqrt{x}} = \int_{k}^{9}\frac{dx}{\sqrt{x}}[/tex]

Do you follow?

Gauss177 said:
2. Let f be a function given by f(x)=x^3 - 6x^2 + p, where p is an arbitrary constant.

a) Write an expression for f '(x) and use it to find the relative maximum and minimum values of f in terms of p. Show the analysis that leads to your conclusion.
My thoughts: f '(x) = 3x^2 - 12x. Setting this equal to 0 and solving for x gives: x=0, 4. These are critical points, and I know that to find max/min points I need to look at second derivative. But here it says just use first derivative? I'm also not sure what to write for analysis part.
Consider the gradient either side of your local extrema, more specifically look at the sign of the gradient (i.e. +ve or -ve). What can this tell you about the local extrema?
Gauss177 said:
b) For what values of the constant p does f have 3 distinct real roots?
I don't have any thoughts on this one :frown:
Factorise the cubic as see where it takes you (take a factor of x2 out)...
Gauss177 said:
c) Find the value of p such that the average value of f over the closed interval [-1, 2] is 1.
Do you know the expression for the average of a function?
 
Last edited:
  • #3
Gauss177 said:
Sorry for so many questions, I'm just trying to understand everything before my test coming up soon.

1. Let R be the region in the first quadrant under the graph of y=1/sqrt(x) for
4<=x<=9.

a) If the line x=k divides the region R into two regions of equal area, what is the value of k?
My thoughts: For this one, I know how to find the area of R, but I don't know how to start finding the line that divides the region into multiple parts of equal area.

We want two equal areas, so perform the following integral.
[tex]
\int_4^k\frac{dx}{\sqrt x}=\int_k^9\frac{dx}{\sqrt x} [/tex]


2. Let f be a function given by f(x)=x^3 - 6x^2 + p, where p is an arbitrary constant.

a) Write an expression for f '(x) and use it to find the relative maximum and minimum values of f in terms of p. Show the analysis that leads to your conclusion.
My thoughts: f '(x) = 3x^2 - 12x. Setting this equal to 0 and solving for x gives: x=0, 4. These are critical points, and I know that to find max/min points I need to look at second derivative. But here it says just use first derivative? I'm also not sure what to write for analysis part.

You need to take the second derivative in order to classify the critical points.

b) For what values of the constant p does f have 3 distinct real roots?
I don't have any thoughts on this one :frown:

Look at the discriminant: http://en.wikipedia.org/wiki/Cubic_equation under the section "the nature of the roots"

c) Find the value of p such that the average value of f over the closed interval [-1, 2] is 1.

Any thoughts?

edit: late
 
  • #4
cristo said:
You need to take the second derivative in order to classify the critical points.
I beg to differ there. Once can classify local extrema by examining the sign of the first derivative either side of the critical point.
 
  • #5
Hootenanny said:
I beg to differ there. Once can classify local extrema by examining the sign of the first derivative either side of the critical point.

Apologies, of course you can. I really meant that, in my opinion, in order to classify the critical points, it is simpler to take the second derivative.
 

1. What is calculus?

Calculus is a branch of mathematics that deals with the study of change and motion. It involves analyzing and solving problems involving rates of change, derivatives, and integrals.

2. What are derivatives?

Derivatives are a fundamental concept in calculus that represent the rate of change of a function. They are used to find the slope of a curve at a specific point and can be thought of as the instantaneous rate of change.

3. How do I find the derivative of a function?

To find the derivative of a function, you can use the power rule, product rule, quotient rule, or chain rule depending on the form of the function. These rules allow you to find the derivative of a function by manipulating its algebraic expression.

4. What is integration?

Integration is the inverse operation of differentiation and is used to find the area under a curve. It involves finding the antiderivative of a function, which is a function whose derivative is equal to the original function.

5. How do I solve optimization problems using calculus?

Optimization problems involve finding the maximum or minimum value of a function. To solve these problems using calculus, you can use the first or second derivative test to determine the critical points of the function and then evaluate them to find the maximum or minimum value.

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