Problems from recent Finals

1 Differentiation Skills: Compute the following derivatives

a)  y = 5 (Sqrt[x]) - 3 (x^2),  find  (dy/dx).

b)  y  =  (e^x) Cos[x],  find  (dy/dx).

c)  y = Sin[(x^2) + 2 x + 1],  find  (dy/dx).

d)  f[x]  =  (x^2) (e^(3  x)),  find  f'[x].

Compute the following derivatives

a)  y = 3 (x^8) - 2 (x^5) + 4 x + 6

(dy/dx)=

b)  y=x (Sqrt[x])

(dy/dx)=

c)  f[x] = (e^(3 x))

f'[x] =

d)  y=x  Log[x]

(dy/dx)=

e)  f[θ] = Sin[ (θ^2) +  3  θ]

f'[θ]=

Differentiation Skills: Compute the following derivatives

a)   y=x (Sqrt[x]),  find  (dy/dx)=

b)  y  =  (e^x) Sin[x],  find  (dy/dx)=

c)  y = Log[(x^2) + 2 x + 1],  find  (dy/dx)=

d)  (dy/dx) =  y(1 - y),  find the second derivative in terms of  y,  (d/dx) ((dy/dx)) = (((d^2) 
                y)/(dx^2))=

2 Integration Skills: Compute the following:

a)   \(\(\(\ \)\(\(∫\((5\ x\^3 - \@x\%3)\) dx\)\(\ \)\(=\)\)\)\)

b)  (∫_(π/6)^(π/4))Cos[2 θ] dθ =

c)  (∫_0^1) ((x^2)/(1 + 
                      (x^3))) dx=

Compute the following intgrals

a)  ∫(2/(x^3))- (x^(1/3)) dx =

b)  (∫_(π/6)^(π/4))Cos[
                  2 θ] dθ =

c) (∫_0^1) ((x^2)/(1 + 
                      (x^3))) dx=

c) ∫x (e^x) dx =

d)  (∫_1^e)x Log[x] dx

e)  ∫ Sin[θ] Cos[θ]dθ =

2 Integration Skills: Compute the following:

a)    ∫(5 (x^3) - 
                2 (Sqrt[ x])) dx =

b)  (∫_a^b) Cos[2  x] dx =

c)  \(∫\(1\/\((2  x + 1)\)\^2\) \(\(dx\)\(\ \)\)\)=

d)  \(\(\(∫x\ \(e\^x\) dx\)\(\ \)\(=\)\)\)

3) Shapes:  Four functions and their derivatives are:

\(TextForm\`f\_1[
        x] = \(x\/\(4 + x\^4\)\ and\ \ \(f\_1'\)[
            x] = \(\(4 - 3\ x\^4\)\/\((4 + x\^4)\)\^2\ \ \ and\ \ \ \(\(f\_1'\)'\)[x] = \(4\ x\^3\ \((3\ x\^4 - 20)\)\)\/\((4 + x\^4)\)\^3\)\)\)

\(TextForm\`f\_2[
        x] = \(x\^2\/\(4 + x\^4\)\ \ and\ \ \(f\_2'\)[
            x] = \(\(2\ x\ \((2 - x\^2)\)\ \((2 + x\^2)\)\)\/\((4 + x\^4)\)\^2\ \ and\ \ \(\(f\_2'\)'\)[
              x] = \(2\ \((16 - 48\ x\^4 + 3\ x\^8)\)\)\/\((4 + x\^4)\)\^3\)\)\)

\(TextForm\`f\_3[
        x] = \(x\^3\/\(4 + x\^4\)\ \ \ and\ \ \(f\_3'\)[
            x] = \(\(x\^2\ \((12 - x\^4)\)\)\/\((4 + x\^4)\)\^2\ \ \ and\ \ \(\(f\_3'\)'\)[
              x] = \(2\ x\ \((48 - 48\ x\^4 + x\^8)\)\)\/\((4 + x\^4)\)\^3\)\)\)

\(TextForm\`f\_4[
        x] = \(x\^4\/\(4 + x\^4\)\ \ and\ \ \(f\_4'\)[
            x] = \(\(16\ x\^3\)\/\((4 + x\^4)\)\^2\ \ and\ \ \(\(f\_4'\)'\)[
              x] = \(16\ x\^2\ \((12 - 5\ x\^4)\)\)\/\((4 + x\^4)\)\^3\)\)\)

a = 4; f1[x_] = x/(a + x^4)
f2[x_] = x^2/(a + x^4)
f3[x_] = x^3/(a + x^4)
f4[x_] = x^4 /(a + x^4)

f1'[x] // Together

Factor[f1'[x]]
Factor[f2'[x]]
Factor[f3'[x]]
Factor[f4'[x]]

f1''[x] // Together

Factor[f1''[x]]
Factor[f2''[x]]
Factor[f3''[x]]
Factor[f4''[x]]

\(
\(grf1 = 
        Plot[f1[x], {x, \(-4\), 4}, PlotRange -> {\(-1\)/2, 1}, 
          AxesOrigin -> {0, 0}, AxesLabel -> {"x", "y"}, 
          PlotLabel -> "Graph d             "];\)\)
grf2 = 
        Plot[f2[x], {x, -4, 4}, PlotRange -> {-1/2, 1}, 
          AxesOrigin -> {0, 0}, AxesLabel -> {"x", "y"}, 
          PlotLabel -> "Graph b             "];
grf3 = 
        Plot[f3[x], {x, -4, 4}, PlotRange -> {-1/2, 1}, 
          AxesOrigin -> {0, 0}, AxesLabel -> {"x", "y"}, 
          PlotLabel -> "Graph a             "];
grf4 = 
        Plot[f4[x], {x, -4, 4}, PlotRange -> {-1/2, 1}, 
          AxesOrigin -> {0, 0}, AxesLabel -> {"x", "y"}, 
          PlotLabel -> "Graph c             "];
Show[GraphicsArray[{{grf3, grf2}, {grf4, grf1}}]];

a)  Which functions and which graphs have a positive derivative at  x = 0?  
Function \(\(f\_?\)[x]\)    and Graph ?

b)  Which functions and which graphs have a positive derivative for all  x > 0?   
Function \(\(f\_?\)[x]\)    and Graph ?

c)  Which functions and which graphs have a positive derivative at  x = -4?  
Function \(\(f\_?\)[x]\)    and Graph ?

d)  Which functions and which graphs have a negative second derivative for x<0, x≈0, and positive second derivative for x>0, x≈0,?  
Function \(\(f\_?\)[x]\)    and Graph ?

e)  Write "(Subscript[f, 1])[x]"  on its graph, "(Subscript[f, 2])[x]'" on its graph,"(Subscript[f, 3])[x]"  on its graph, and "(Subscript[f, 4])[x]"  on its graph.

4) MAX-min:

a) Given that f[x] = (((1 - (x^2)))^2) + 2 (c^2) (x^2), show that  f ' [x]  =  4 x ((x^2) - 1 + (c^2))  and find all values where 
 f ' [x] = 0 .  (Express your answer in terms of  c.)

b)  Find the maximum and minimum of the function  f[x] = (((1 - (x^2)))^2) + 2 (c^2) (x^2)  for  -1 ≤ x ≤ +1, in terms of the parameter  c  when  0 < c < (1/(Sqrt[2])).

c)  Find the maximum and minimum of the function  f[x] = (((1 - (x^2)))^2) + 2 (c^2) (x^2)  for  -1 ≤ x ≤ +1, in terms of the parameter  c  when  (1/(Sqrt[2])) <  c  <  1.

d)  Find the maximum and minimum of the function  f[x] = (((1 - (x^2)))^2) + 2 (c^2) (x^2)  for  -1 ≤ x ≤ +1, in terms of the parameter  c  when   c  >  1.

5)  Speak the Language

FACT 1:  Radioactive mathonium atoms decay randomly with a constant probability and when they do, they emit alpha particles and split into harmless inert englishium and businessium atoms.  The more mathonium there is, the more alpha particles will be emitted in a given second.  The more alpha particles observed, the less mathonium is left because the loss of mathonium is proportional to the emitted alpha particles.

FACT 2:  At the beginning of the semester I had 45 grams of mathonium.

FACT 3: At mid-term (after 8 weeks), I had only 15 grams left.

a)  Let  m = the amount of mathonium at any given time in grams.  
Let  t = the elapsed time in weeks with  t=0 at the beginning of the semester.  Use FACT 1 to give and expression for the rate of change of  m  with respect to  t.

b)  Is the rate in your answer to part  (a)  positive or negative?  How is this represented in the constants of your expression?

c)  Combine your answer to part (a) and FACT 2  to express the amount of mathonium as an initial value problem.

d)  How much mathonium will I have left after finals (16 weeks?)  Explain your computation.

3 Knee Jerk

A function of time,  y =  y[t]  satisfies:  y[0]  =  3   & (dy/dt) = -(1/2) y

Give the exact value of  y[4] = ? and explain your formula.  (Note:  Log[2]  is exact, 0.693147, is approxiamte.  Sin[1]  is exact, 0.841471,  is approximate.)

5)  Max-min

What are the largest and smallest values that the function f[x] = ((x^2)/(3 + (x^2)))could attain for values of  x ≥ 0 ?  You are given the derivative  f'[
          x]  =  ((6 x)/(((3 + (x^2)))^(2 )))

Does  f[x]  actually attain these values or are they limiting cases?  Prove your assertions for example by giving a sketch of the graph for  x  in [0,∞) with sufficient supporting facts.

Max-min

What are the largest and smallest values that the function f[x] = ((x^2)/(3 + (x^2)))could attain for values of  x ≥ 0 ?  You are given the derivative  f'[
          x]  =  ((6 x)/(((3 + (x^2)))^(2 )))

Does  f[x]  actually attain these values or are they limiting cases?  Prove your assertions for example by giving a sketch of the graph for  x  in [0,∞) with sufficient supporting facts.

More Max-min

Find the max and min of the function  f[x] = (x^3) - 12 x + 5  for  x  in the interval  [-3,5].  Does  f[x]  have both a max and min?  Explain your reasoning carefully.

Graphing

Four functions and their derivatives are:

(Subscript[f, 1])[
        x] = x (e^x)  and  (Subscript[f, 1])'[
            x] = (1 + x) (e^x)  and   (Subscript[f, 1])''[
              x] = (2 + x) (e^x)

(Subscript[f, 2])[
        x] = (x^2) (e^x)  and  (Subscript[f, 2])'[
            x] = x (x + 
                2) (e^x)   and  (Subscript[f, 2])''[
              x] = (2 + 4 x + (x^2)) (e^x)

\(TextForm\`f\_3[
        x] = \(x\ e\^\(x\^2\)\ \ and\ \ \(f\_3'\)[
            x] = \(\((1 + 
                2\ x\^2)\)\ e\^\(x\^2\)\ \ \ and\ \ \(\(f\_3'\)'\)[x] = 
          2\ x\ \((3 + 2\ x\^2)\)\ e\^\(x\^2\)\)\)\)

\(TextForm\`f\_4[
        x] = \(e\^\(x\^3\)\ \ and\ \ \(f\_4'\)[
            x] = \(3\ \(x\^2\) e\^\(x\^3\)\ \ \ and\ \ \(\(f\_4'\)'\)[x] = 
          3\ x\ \((2 + 3\ x\^3)\) e\^\(x\^3\)\)\)\)

Match the  4  graphs below with the  4  formulas above by writing  "\(TextForm\`f\_3[x] \*"""\), etc. on the proper graph.

f1[x_] = x (E^x)
f2[x_] = (x^2) (E^x)
f3[x_] = x (E^((x^2)))
f4[x_] = (E^((x^3)))

Factor[f1'[x]]
Factor[f2'[x]]
Factor[f3'[x]]
Factor[f4'[x]]

Factor[f1''[x]]
Factor[f2''[x]]
Factor[f3''[x]]
Factor[f4''[x]]

\(
\(grf1 = 
        Plot[f1[x], {x, \(-2\), 3/2}, PlotRange -> {\(-2\), 5}, 
          AxesOrigin -> {0, 0}, AxesLabel -> {"x", "y"}, 
          Ticks -> None, PlotLabel -> "Graph d        "];\)\)
grf2 = 
        Plot[f2[x], {x, -2, 3/2}, PlotRange -> {-2, 5}, 
          AxesOrigin -> {0, 0}, AxesLabel -> {"x", "y"}, 
          Ticks -> None, PlotLabel -> "Graph b        "];
grf3 = 
        Plot[f3[x], {x, -2, 3/2}, PlotRange -> {-2, 5}, 
          AxesOrigin -> {0, 0}, AxesLabel -> {"x", "y"}, 
          Ticks -> None, PlotLabel -> "Graph c        "];
grf4 = 
        Plot[f4[x], {x, -2, 3/2}, PlotRange -> {-2, 5}, 
          AxesOrigin -> {0, 0}, AxesLabel -> {"x", "y"}, 
          Ticks -> None, PlotLabel -> "Graph a        "];
Show[GraphicsArray[{{grf4, grf2}, {grf3, grf1}}]];

The graph of  y =  ((x (((x - 2))^3) (x-6))^2)= -288 x + 528 (x^2) - 368 (x^3) + 120 (x^4) - 18 (x^5) + 
        (x^6)  is shown below.  The x-values indicated at the bottom are the ONLY places where either  (dy/dx)= 0 or  (((d^2) y)/(dx^2)) = 0.   (x = 0.44, 0.88, 2.0, 3.6, 5.5, 6.0)  Make a complete shape table for this function showing where y  is increasing and decreasing and where  y = f[x]  is convex up or down.

Clear[x, y];
y[x_] = x (((x - 6))^2) (((x - 2))^3)
Expand[y[x]]
NSolve[(∂_x) y[x] == 0]
\(NSolve[∂\_{x, 2}y[x] == 0, x]\)
curve = 
        Plot[y[x], {x, - .2, 6.5}, 
          AxesLabel -> {"x", "y"}];

Show[curve, 
        Graphics[{Line[{{ .438, y[ .438]}, { .438, -55}}], 
            Line[{{ .88, y[ .88]}, { .88, -55}}], 
            Line[{{2, y[2]}, {2, -55}}], 
            Line[{{3.65, y[3.65]}, {3.65, -55}}], 
            Line[{{5.47, y[5.47]}, {5.47, -55}}], 
            Line[{{4.56, y[4.56]}, {4.56, -55}}], 
            Line[{{6, y[6]}, {6, -55}}]}], 
        Graphics[{Text[".44", { .438, -65}], 
            Text[".88", { .88, -65}], Text[2. , {2, -65}], 
            Text[3.6, {3.65, -65}], Text[4.6, {4.56, -65}], 
            Text[5.5, {5.47, -65}], 
            Text[6.0, {6, -65}]}]];

Differential Equation describing something

Fact 1:  A certain chemical reaction proceeds at a rate proportional to the concentration of a compound  C.  The reaction is faster when the concentration of compound  C  is higher and slower as the concentration of  C  is lower and the relation between reaction rate and concentration is linear for this particular compound.

Fact 2:  This reaction uses the compound  C  up, so the faster the reaction is going, the more compound  C  is being used per second.

Fact 3:  The reaction begins with a concentration of  2.0 millimolars at time  0  and 3 seconds later, the concentation is  1.5 millimolars.

Let
	c = the concentration of compound  C  in millimolars
	t = time in seconds beginning with  t=0 at the start of the reaction

a)  Express Fact 1 by giving an expression for the reaction rate = ???

b)  Use Fact 2 to describe the reaction rate in terms of the derivative  (dc/dt) and combine this with part  (a)  to give a differential equation for the concentration of  c.

(dc/dt)  =  ??

c)  Use Fact 3  and  part  (b)  to find a formula c = c[t] = ???  valid for all   t ≥0.

c[t_] =  2 E^(-k t)
\(		Solve[c[3] == 3/2, k]\)