Critical points:
Find x-values which can make your first derivative equal zero or cause it to become undefined. The reason this is important is because the first derivative gives us slope, and we need to find points where the slope is 0 (horizontal line) because this means the graph must be in the middle of changing from a negative to positive or positive to negative slope, etc. at that point. The reason it is important to also find places where the slope may not exist is to find bents, sharp turns, vertical tangent lines (slope = infinity), discontinuities, etc. (whatever could cause a derivative to not exist).
The first derivative tells us about the slope of the function, specifically is the function increasing or decreasing?
If f’(x) is negative, f(x) is decreasing
If f’(x) is positive, f(x) is increasing.
The second derivative tells us about the concavity, specifically is the function concave up or concave down?
If f”(x) is negative, f(x) is concave down.
If f”(x) is positive, f(x) is concave up.
Finding Local Maximums and Minimums (Relative Extrema):
The first derivative can also help to find if a critical point is a local minimum or maximum (test intervals).
If f’(x) changes from positive (left of the critical point) to negative (right of the critical point) the critical point is a maximum.
If f’(x) changes from negative (left of the critical point) to positive (right of the critical point) the critical point is a minimum.
However, if the second derivative can be easily obtained, we can use the second derivative test to determine if a critical point is a local minimum or maximum.
Once you get the first derivative and critical points, you can plug in those critical points into the second derivative and check whether the value you get when you solve is positive or negative.
If f”(x) is negative, then the critical point would be a local maximum (concave down).
If f”(x) is positive, then the critical point would be a local minimum (concave up).
Inflection points: x-values that cause the second derivative to equal zero or cause it to become undefined. These are points where the graph is in the middle of changing from concave up to concave down or concave down to concave up.
Test Intervals:
Intervals can be used to find whether critical points are local maximums or minimums, as well as help us have a brief understanding of how our function looks (where it’s increasing and where it’s decreasing).
We make intervals to examine what’s happening to the left of a critical point and to the right of the critical point. If we plug in any value that exists within these intervals into the first derivative, we simply examine whether the value is positive or negative. Since the first derivative gives us the slope at a point, if it is negative, we know the graph would be decreasing within that interval until it reaches the next critical point. If it was positive, then it would increase within that interval until it reaches the next critical point.
Example, if we let g and k be two critical points where k > g, and the domain of the original function is all real numbers, then the intervals would be like so:
- ∞ < g (to the left of g) | g < k (between g and k) | k < ∞ (to the right of k)
Related rates word problems:
1.) Draw a diagram if possible.
2.) Label any constants and variables.
3.) Identify rates [(dn/dt), change in a variable (n) with respect to time (t)], and values of variables. You can use a formula that relates all variables to solve for unknown variables.
4.) Without plugging in any values, take the derivative of the entire equation or formula that relates these variables. Make sure to take the derivative of both sides with respect to time (d/dt).
5.) Substitute known variables and rates into the derived equation, (with the values from step 3).
You should be left with one thing to solve for, solve for it!
How to approach optimization word problems:
1.) Draw a diagram/ sketch/ graph if possible.
2.) Label any constants and use variables for things that’ll help in maximization or minimization.
3.) Identify formula(s) that can be used to relate the constants and variables. (Usually, the actual question part of the problem gives you a hint at what type of formula to use ex. find maximum area, volume or find minimum cost, distance).
4.) Reduce it down to one independent variable using the formula(s). (I put formulaS because you may need another formula that relates the variables, especially when you look at the quantities that are given in the question (ex. perimeter (given)... find max/min area).
5.) Take the derivative of the final reduced equation with respect to the one independent variable.
Find critical points. This is where you can use one of the two approaches:
If the second derivative is easy to get, then simply plug the critical point into the second derivative and identify whether the answer is a negative or positive number. This will tell you whether the function is concave down (in which case the critical point is a local max.) or concave up (critical point is a local min.) respectively.
If the second derivative isn’t easy to get, then make test intervals and plug in points that exist in those intervals into the first derivative to examine the left and right sides of the critical point. If the left side is negative and right side is positive, then the critical point is a local min., and vice versa (left is positive and right is negative, then critical point is local max.).
Once you identify a maximum or minimum (whatever you were solving for), you can use it to solve any other unknowns if you need it (ex. other side lengths).
***Velocity is the first derivative (rate of change in position (displacement)); acceleration is the second derivative (rate of change in velocity).
Rectilinear motion: movement of a particle along a line, where velocity is positive when it moves forward, 0 when it stops and negative when it moves backwards.
Mean Value Theorem: The slope of the secant line that goes through two x-values that make an interval should be parallel to the slope of a tangent line somewhere in between that interval at x = c (assuming function is both differentiable and continuous everywhere in that interval).
L’Hopital’s Rule: For indeterminate forms where the fraction has a numerator and a denominator like any of the following:
- 0/0
- infinity/infinity
Use L’Hopital’s rule.
L’Hopital’s Rule: Derive the numerator and denominator (keep it in fractional form) and then plug in the value... limit as x approaches (value).
No comments:
Post a Comment