IndietroBusiness Calculus: Critical Points, Derivative Tests, and Asymptotes
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Q1. Find the x-coordinates of any critical points for the function . You only need to determine the x-coordinates, not test them.
Background
Topic: Finding Critical Points
This question is testing your ability to find critical points of a function, which are candidates for local maxima or minima. Critical points occur where the first derivative is zero or undefined.
Key Terms and Formulas
Critical Point: A value of where or does not exist.
First Derivative: gives the slope of the tangent line to the function at any point .
Step-by-Step Guidance
Find the first derivative of . That is, compute .
Set to find where the slope of the function is zero.
Solve the resulting equation for to find the critical point(s).
Check if there are any values where does not exist. (For this polynomial, the derivative exists everywhere.)
Try solving on your own before revealing the answer!
Final Answer:
The critical point(s) occur at .
We found , set it to zero, and solved for .
Q2. The function has a critical point at . Use the FIRST Derivative Test to determine if the critical point is a maximum, minimum, or neither.
Background
Topic: First Derivative Test for Local Extrema
This question is about using the first derivative test to classify a critical point as a local maximum, minimum, or neither. The test involves checking the sign of the derivative before and after the critical point.
Key Terms and Formulas
First Derivative Test: If changes from negative to positive at , has a local minimum at . If $f'(x)$ changes from positive to negative, $f$ has a local maximum at $c$.
Critical Point: Where or is undefined.
Step-by-Step Guidance
Find the first derivative of the function.
Evaluate at values just less than and just greater than (for example, at and ).
Determine the sign of on each side of .
Use the First Derivative Test: If the sign changes from negative to positive, it's a minimum; if from positive to negative, it's a maximum.
Try solving on your own before revealing the answer!
Final Answer:
The critical point at is a minimum.
The derivative changes from negative to positive as increases through 4.
Q3. The function has a critical point at . Use the SECOND Derivative Test to determine if the critical point is a maximum or a minimum, or indicate if the test is inconclusive.
Background
Topic: Second Derivative Test for Local Extrema
This question is about using the second derivative test to classify a critical point. The test involves evaluating the second derivative at the critical point.
Key Terms and Formulas
Second Derivative Test: If , has a local minimum at . If , $f$ has a local maximum at $x = c$. If , the test is inconclusive.
Second Derivative: is the derivative of .
Step-by-Step Guidance
Find the first derivative of the function.
Find the second derivative .
Evaluate at .
Use the sign of to determine if the critical point is a maximum, minimum, or if the test is inconclusive.
Try solving on your own before revealing the answer!
Final Answer:
The critical point at is a maximum.
Since , the function is concave down at this point.
Q4. Does the function have any vertical asymptotes? If so, indicate the position of the asymptote. If not, explain why.
Background
Topic: Vertical Asymptotes of Rational Functions
This question is about identifying vertical asymptotes, which occur where the denominator of a rational function is zero and the numerator is not zero at that point.
Key Terms and Formulas
Vertical Asymptote: A vertical line where the function approaches infinity or negative infinity as approaches .
Rational Function: A function of the form .
Step-by-Step Guidance
Set the denominator equal to zero and solve for .
Check if the numerator is also zero at this value of (if so, it may be a hole instead of an asymptote).
If the denominator is zero and the numerator is not zero, there is a vertical asymptote at that value.
Try solving on your own before revealing the answer!
Final Answer:
There is a vertical asymptote at .
The denominator is zero at , and the numerator is not zero there, so the function has a vertical asymptote at this point.