Calculus I - Applications of the Derivative

28 flashcards · Mathematics · by luckydog789 (harrison holbrook)

1. Absolute Maximum

   Assume a function fff is defined on a set DDD , and x=cx=cx=c is a point in DDD . f(c)f(c)f(c) is an absolute maximum of fff on DDD if f(c)≥f(x)f(c)\ge f(x)f(c)≥f(x) for every xxx in DDD . Undefined points, even if they have a limit, can never be an absolute maximum.

2. Absolute Minimum

   Assume a function fff is defined on a set DDD , and x=cx=cx=c is a point in DDD . f(c)f(c)f(c) is an absolute minimum of fff on DDD if f(c)≤f(x)f(c)\le f(x)f(c)≤f(x) for every xxx in DDD . Undefined points, even if they have a limit, can never be an absolute minimum.

3. Extreme Value Theorem (EVT)

   A function that is continuous on a closed interval is guaranteed to have both an absolute maximum value and an absolute minimum value. However, discontinuous functions, or functions defined on an (semi or fully) open interval may still have absolute extrema.

4. Local Maxima

   Assume x=cx=cx=c is an interior point (i.e. not an endpoint) of some interval III in the domain of fff . f(c)f(c)f(c) is a local maximum value of fff if f(c)≥f(x)f(c)\ge f(x)f(c)≥f(x) for every xxx in III . Stated more simply, f(c)f(c)f(c) is a local maxima if it is greater than all nearby points, but not necessary every point in the interval.

5. Local Minima

   Assume x=cx=cx=c is an interior point (i.e. not an endpoint) of some interval III in the domain of fff . f(c)f(c)f(c) is a local minimum value of fff if f(c)≤f(x)f(c)\le f(x)f(c)≤f(x) for every xxx in III . Stated more simply, f(c)f(c)f(c) is a local minima if it is less than all nearby points, but not necessary every point in the interval.

6. The overlap between maxima and minima values

   There are always at most one absolute maximum and one absolute minimum, but there can be infinity many local minima and maxima. Additionally, local minima and maxima values can be the absolute maximum value and the absolute minimum value respectively.

7. Local Extreme Value Theorem

   If a function fff has a local maxima or minima at a point x=cx=cx=c , then either f′(c)=0f'(c)=0f′(c)=0 or f′(c)f'(c)f′(c) does not exist. (Therefore there exists a critical point at ccc ) (plus the converse of this; if there exists a local maximum or minimum at ccc in a differentiable function, f′(c)=0f'(c)=0f′(c)=0 )

8. Critical Point for Maxima and Minima / Increasing and Decreasing

   An interior point x=cx=cx=c of the domain of fff is a critical point of fff if either f′(c)=0f'(c)=0f′(c)=0 or f′(c)f'(c)f′(c) does not exist. If fff has a local extrema, it must occur at a critical point, but not every critical point is guaranteed to be a local extrema.

9. Steps for finding absolute extrema of a continuous function fff on interval [a,b][a,b][a,b]

   Make a list of all critical points of fff Discard any critical point outside the interval (a,b)(a,b)(a,b) from the list Add the endpoints x=ax=ax=a and x=bx=bx=b to the list Find the yyy -values of fff for all the xxx -values in the list The greatest yyy -value found in Step 4 is the absolute maximum value of fff on [a,b][a,b][a,b] , and the least yyy -value found in Step 4 is the absolute minimum value of fff on [a,b][a,b][a,b]

10. Definition of an increasing or decreasing function

    Suppose a function fff is defined on an interval III . fff is increasing on III if f(x_1)"> f(x2)>f(x1)f(x_2)>f(x_1)f(x2​)>f(x1​) whenever x2x_2x2​ and x1x_1x1​ are in III and x_1"> x2>x1x_2>x_1x2​>x1​ , and fff is decreasing on III if f(x2)<f(x1)f(x_2)<f(x_1)f(x2​)<f(x1​) whenever x2x_2x2​ and x1x_1x1​ are in III and x_1"> x2>x1x_2>x_1x2​>x1​ .

11. Derivative definition of an increasing function

    fff is increasing if every line tangent to fff on an interval III has a positive slope, and therefore 0"> f′(x)>0f'(x)>0f′(x)>0 for all xxx in III . This is a corollary of the MVT.

12. Derivative definition of an decreasing function

    fff is increasing if every line tangent to fff on an interval III has a negative slope, and therefore f′(x)<0f'(x)<0f′(x)<0 for all xxx in III . This is a corollary of the MVT.

13. Test for determining if a function is increasing or decreasing on an interval

    Suppose a function fff is defined on an interval III , and differentiable on all points inside III . If 0"> f′(x)>0f'(x)>0f′(x)>0 for all interior points of III , then fff is increasing on III . If f′(x)<0f'(x)<0f′(x)<0 for all interior points of III , then fff is decreasing on III . This is a corollary of the MVT.

14. Steps for determining on which intervals a function fff is increasing or decreasing

    Find all critical points of fff Make a sign diagram (i.e. the number line with intervals marked of the critical points, with the sign of each interval above) Determine the sign of each interval by determining the sign of fff at some value in the interval Describe the intervals in open notation (ex. (−∞,0)(-\infty ,0)(−∞,0) )

15. The first derivative test

    Used to identify locations/types of local extrema Assume that x=cx=cx=c is a critical point of fff fff is continuous on an interval containing ccc f′f'f′ exists on an interval containing ccc (except possibly at ccc ) There are four possibilities for the sign graph of f′f'f′ near x=cx=cx=c <+++ ccc ---> ⟹ \implies ⟹ f(c)f(c)f(c) is a local maxima <--- ccc +++> ⟹ \implies ⟹ f(c)f(c)f(c) is a local minima <+++ ccc +++> ⟹ \implies ⟹ f(c)f(c)f(c) isn't a local extrema <--- ccc ---> ⟹ \implies ⟹ f(c)…

16. One local extrema theorem

    Suppose fff is continuous on an interval III that contains exactly one local extrema at x=cx=cx=c . This local extrema is an absolute extreme of fff on III (with the type dependent on the type of the local extrema).

17. Concavity

    suppose a function fff is twice differentiable on an open interval III . If f′f'f′ is increasing on III , then 0"> f′′>0f''>0f′′>0 on III and fff is concave up on III If f′f'f′ is decreasing on III , then f′′<0f''<0f′′<0 on III and fff is concave down on III

18. Inflection point

    Suppose a function fff is twice differentiable on an open interval III . If fff is continuous at a point ccc in III and fff changes concavity at ccc , then fff has an inflection point at ccc , and f′′f''f′′ must change sign as x increases through ccc (but if f′′f''f′′ changes sign, it is not always an inflection point).

19. Steps to determine intervals of concavity and inflection points of f(x)f(x)f(x)

    Calculate f′(x)f'(x)f′(x) then f′′(x)f''(x)f′′(x) Determine all points where f′′=0f''=0f′′=0 or f′′f''f′′ does not exist (almost think of them as the critical points of f′(x)f'(x)f′(x) ) Make a sign graph of f′′f''f′′ using the possible inflection points from Step 2 to split the domain of fff into intervals Test each interval to determine its sign Analyze the sign graph If 0"> f′′>0f''>0f′′>0 on an interval, fff is concave up on the interval if f′′<0f''<0f′′<0 on an interval, fff is concave dow…

20. The second derivative test

    Assume f′′f''f′′ is continuous on an open interval containing x=cx=cx=c , with f′(c)=0f'(c)=0f′(c)=0 (we know ccc is a critical point) If 0"> f′′(c)>0f''(c)>0f′′(c)>0 , then fff has a local minimum at ccc If f′′(c)<0f''(c)<0f′′(c)<0 , then fff has a local maximum at ccc If f′′(c)=0f''(c)=0f′′(c)=0 , then the test is inconclusive, and the first derivative test can be used to determine whether there is a local maxima, minima, or neither at x=cx=cx=c

21. Guidelines for graphing a function fff

    Identify the domain of fff , or intervals of interest Consider symmetry (even, odd, or neither) Find f′(x)f'(x)f′(x) and f′′(x)f''(x)f′′(x) Find all critical points and possible inflection points Find intervals on which fff is increasing or decreasing, and intervals on which fff is concave up or concave down Identify local extrema, inflection points, and horizontal tangents (using the sign graphs from the previous step) Locate asymptotes and identify end behavior Find the xxx and yyy intercepts…

22. Optimization problem

    Goal: Find an optimized solution to a problem, such as the minimum or maximum of a given function, called the objective function Optimization problems introduce constraints which can be used to define relationships between variables, allowing the objective function to be re-written in terms of one variable, allowing it to be differentiated

23. Guidelines for optimizations problems

    Read the problem and organize information, initially in a picture if needed before being in an equation Identify the function being optimized Identify all constraints and describe them in terms of the variables of the problem Use the constraints to re-write the objective function in terms of only one variable Identify the interval of interest for the problem Use calculus to optimize the problem (often by finding a min or max of the objective function, remember to consider endpoints, not just cr…

24. Linear approximation

    An approximation of a function at aaa at values close to aaa (Almost image how zooming into the curve of x2x^2x2 almost looks like a straight line at good enough zoom) Suppose a linear approximation LLL of function fff near x≅ax\cong ax≅a L(x)=f(a)+f′(a)(x−a)L(x)=f(a)+f'(a)(x-a)L(x)=f(a)+f′(a)(x−a)

25. Differentials

    The change in a linear approximation for a small change in the independent variable Δx=\Delta x =Δx= some change in xxx Δy=\Delta y =Δy= some change in yyy from xxx to x+Δxx + \Delta xx+Δx dx=Δx≠0dx=\Delta x \neq 0dx=Δx=0 dy=f′(x)dx=dy=f'(x)dx=dy=f′(x)dx= change in height of the tangent of the graph from xxx to xxx + Δx\Delta xΔx

26. Approximating Error

    If fff is a function and Δf\Delta fΔf is the change in fff , then the Relative error at x0x_0x0​ of fff is ∣Δf∣f(x0)\frac{|\Delta f|}{f(x_0)}f(x0​)∣Δf∣​ The percent error at x0x_0x0​ of fff is ∣Δf∣f(x0)\frac{|\Delta f|}{f(x_0)}f(x0​)∣Δf∣​ ⋅100\cdot 100⋅100

27. Newton's Method for approximating zeros

    Suppose a function fff is defined on a closed interval [a,b][a,b][a,b] and its first derivative f′(x)f'(x)f′(x) is continuous and not equal to zero on the open interval (a,b)(a,b)(a,b) . If x=c1x=c_1x=c1​ is a sufficiently close first approximation to a real zero for the function fff in (a,b)(a,b)(a,b) , then the formula c2=c1−f(c1)f′(c1)c_2=c_1-\frac{f(c_1)}{f'(c_1)}c2​=c1​−f′(c1​)f(c1​)​ gives a second approximation to the zero. The nnn th approximation to the zero is given by cn=cn−1−f(cn−1)…

28. Related rate problems

    Write an equation involving time-dependent variables From geometric/physical conditions Differentiate the equation (via implicit differentiation) with respect to time and obtain a new equation describing the relationship Use given values in their relationship to solve the question