Calculus/Differentiation/Applications of Derivatives/Exercises

< Calculus‎ | Differentiation

Relative ExtremaEdit

Find the relative maximum(s) and minimum(s), if any, of the following functions.

1.  f(x) = \frac{x}{x+1} \,


2.  f(x) = (x-1)^{2/3} \,

Minimum at the point (1,0)

3.  f(x) = x^2 + \frac{2}{x} \,

Relative minimum at x=1

4.  f(s) = \frac{s}{1+s^2} \,

Relative minimum at s=-1
Relative maximum at s=1

5.  f(x) =  x^2 - 4x + 9 \,

Relative minimum at x=2

6.  f(x) = \frac{x^2 + x +1}{x^2 -x +1} \,

Relative minimum at x=-1
Relative maximum at x=1


Range of FunctionEdit

7. Show that the expression x+ 1/x cannot take on any value strictly between 2 and -2.

1-\frac{1}{x^{2}}=0\implies x=\pm1
Since f^{\prime\prime}(-1) is negative, x=-1 corresponds to a relative maximum.
\lim\limits _{x\to-\infty}f(x)=-\infty
For x<-1, f'(x) is positive, which means that the function is increasing. Coming from very negative x-values, f increases from a very negative value to reach a relative maximum of -2 at x=-1.
For -1<x<1, f'(x) is negative, which means that the function is decreasing.
Since f^{\prime\prime}(1) is positive, x=1 corresponds to a relative minimum.
Between [-1,0) the function decreases from -2 to -\infty, then jumps to +\infty and decreases until it reaches a relative minimum of 2 at x=1.
For x>1, f'(x) is positive, so the function increases from a minimum of 2.
The above analysis shows that there is a gap in the function's range between -2 and 2.

Absolute ExtremaEdit

Determine the absolute maximum and minimum of the following functions on the given domain

8.  f(x) = \frac{1}{3}x^3 - \frac{1}{2}x^2 + 1 on [0,3]

Maximum at (3,\frac{11}{2}); minimum at (1,\frac{5}{6})

9.  f(x) = (\frac{4}{3}x^2 -1)x on [-\frac{1}{2},2]

Maximum at (2,\frac{26}{3}); minimum at (\frac{1}{2},-\frac{1}{3})


Determine Intervals of ChangeEdit

Find the intervals where the following functions are increasing or decreasing

10. f(x)=10-6x-2x^2

Increasing on (-\infty,-\frac{3}{2}); decreasing on (-\frac{3}{2},+\infty)

11. f(x)=2x^3-12x^2+18x+15

Decreasing on (1,3); increasing elsewhere

12. f(x)=5+36x+3x^2-2x^3

Increasing on (-2,3); decreasing elsewhere

13. f(x)=8+36x+3x^2-2x^3

Increasing on (-2,3); decreasing elsewhere

14. f(x)=5x^3-15x^2-120x+3

Decreasing on (-2,4); increasing elsewhere

15. f(x)=x^3-6x^2-36x+2

Decreasing on (-2,6); increasing elsewhere


Determine Intervals of ConcavityEdit

Find the intervals where the following functions are concave up or concave down

16. f(x)=10-6x-2x^2

Concave down everywhere

17. f(x)=2x^3-12x^2+18x+15

Concave down on (-\infty,2); concave up on (2,+\infty)

18. f(x)=5+36x+3x^2-2x^3

Concave up on (-\infty,\frac{1}{2}); concave down on (\frac{1}{2},+\infty)

19. f(x)=8+36x+3x^2-2x^3

Concave up on (-\infty,\frac{1}{2}); concave down on (\frac{1}{2},+\infty)

20. f(x)=5x^3-15x^2-120x+3

Concave down on (-\infty,1); concave up on (1,+\infty)

21. f(x)=x^3-6x^2-36x+2

Concave down on (-\infty,2); concave up on (2,+\infty)


Word ProblemsEdit

22. You peer around a corner. A velociraptor 64 meters away spots you. You run away at a speed of 6 meters per second. The raptor chases, running towards the corner you just left at a speed of 4t meters per second (time t measured in seconds after spotting). After you have run 4 seconds the raptor is 32 meters from the corner. At this time, how fast is death approaching your soon to be mangled flesh? That is, what is the rate of change in the distance between you and the raptor?

10 m/s

23. Two bicycles leave an intersection at the same time. One heads north going 12 mph and the other heads east going 5 mph. How fast are the bikes getting away from each other after one hour?

13 mph

24. You're making a can of volume 200 m^3 with a gold side and silver top/bottom. Say gold costs 10 dollars per m^2 and silver costs 1 dollar per m^2. What's the minimum cost of such a can?



Graphing FunctionsEdit

For each of the following, graph a function that abides by the provided characteristics

25. f(1)= f(-2) = 0, \; \lim_{x\to \infty} f(x) = \lim_{x\to -\infty} f(x) = 0, \; \mbox{ vertical asymptote at } x=-3, \; f'(x)>0 \mbox{ on } (0,2),  f'(x)<0 \mbox{ on } (-\infty,-3)\cup(-3,0)\cup(2,\infty),\; f''(x)>0 \mbox{ on } (-3,1)\cup (3,\infty),\; f''(x)<0 \mbox{ on } (-\infty,-3)\cup(1,3).
There are many functions that satisfy all the conditions. Here is one example:
Calculus graphing exercise 1.png
26. f \mbox{ has domain } [-1,1], \; f(-1) = -1, \; f(-\frac{1}{2}) = -2,\; f'(-\frac{1}{2}) = 0,\; f''(x)>0 \mbox{ on } (-1,1)
There are many functions that satisfy all the conditions. Here is one example:
Calculus graphing exercise 2.png


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