22M:22 Calculus & Modeling 1
Review Problems for Exam 2

1 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.)

Enter the next program:

Clear[y, t];
y[t_] = 3 (E^(-t/2))
y[4]

or, if you don't know the essential fact that the solution to y[0] = 3  & (dy/dt) = -y/2 is  y[t] = 3 (e^(-t/2)), enter the next program:

Clear[y, t];
soln = 
        DSolve[{y[0] == 3, y'[t] == -y[t]/2}, {y[t]}, 
          t];
y[t_] = y[t] /. soln[[1]]
y[4]

2  Max min

Find the max and min values of the function  f[x]  =  (x^3) - 12  x for values of  x   satisfying  -1 ≤ x ≤  3, if the max and min exist or say why they don't if they don't exist.

The Extreme Val;ue Theorem applies because  f[x] is smooth on the compact domain, -1≤x≤3.  Enter the next program.  It defines f[x], computes the derivative and solves for the critical points.

f[x_] = (x^3) - 12 x;
f'[x]
Solve[f'[x] == 0]

Enter the next program.  It defines computes  f[x]  at the endpoints and interior critical points.  The max and min are amongst these by the Interios Critial Points Theorem and Extreme Value Theorem.

f[-1]
f[2]
f[3]

Max value is 11 at f[-1].  Min value is -16 at  x=2.

2TOO Max-min

Find the max and min of the function  f[x] = (x^3) - 12 x + 5  for  x  in the interval  [-1,3].  Does  f[x]  have both a max and min?  Explain your reasoning carefully.

f[x_] = (x^3) - 12 x + 5;
f'[x]
Solve[f'[x] == 0]

f[-1]
f[2]
f[3]

Max value is 16 at f[-1].  Min value is -11 at  x=2.  The Extreme Value Theorem applies because  f[x] is smooth on the interval and [-1,3] is compact.  Hence (by the EVT) f[x] attains both a max and min.  The critical point theorem says that if the max or min is interior, it must be at  x=2.  The endpoints  x=-1 and x=3 are also possible locations.

3 Max min

Find the max and min of the function  f[x] = (1/(((1 - (x^2)))^2))  for all possible  x-values, if the max and min exist, or say why they don't if they don't exist.  (Note: f'[x] = ((4 x)/(((1 - (x^2)))^3))  and  f''[
          x] = ((4 (1 + 5 (x^2)))/(((-1 + (x^2)))^4)).)  Does the function have a limiting "maxima" or "minima", that is, an "infimum" or "supremum" that bounds the function, but that is never exactly attained?

Clear[f, x];
f[x_] = 1/(((1 - (x^2)))^2);
f'[x]

Solve[f'[x] == 0]

f'[-1]

f'[1]

Limit[f[x], x -> +1]

Limit[f[x], x -> -1]

f'[-2]

f'[-1/2]

f'[1/2]

f'[2]

Limit[f[x], x -> +∞]

Limit[f[x], x -> -∞]

Plot[f[x], {x, -2, 2}, PlotRange -> {0, 5}, 
      PlotStyle -> RGBColor[1, 0, 0]]

This function does not have a maximum value.  It tends to +∞  as  x ->±1.  It has a "local minimum" at x=0, but this is of no consequence.  f[x]  thend to zero as  x ->±∞, but it never achieves the value zero, so it does not "have" a minimum.  (It has "infimum" zero as the limit.)

4 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 )))

Clear[f, x];
f[x_] = (x^2)/(3 + x^2);
Together[f'[x]]

Solve[f'[x] == 0]

f'[-1]

f'[1]

The slope table is down, over, up, so x=0 is a minimum, but there is no maximum ATTAINED.

Limit[f[x], x -> ∞]

The supremum of the function is  1, but it is not attained.

Plot[f[x], {x, -10, 10}, PlotRange -> {0, 1}, 
        PlotStyle -> RGBColor[1, 0, 0]];

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.

5 Max-min

Find the dimensions of the largest right circular cylinder that can be inscribed n a sphere of radius  25 cm.  A cut-away view of one possible cylinder is shown on the left with a front-view diagram in the middle.  Another possible cylinder on the right.  The volume of a cylinder of radius  r  and height  h  is  V = π (r^2) h.  Notice the relationship between the radius of the cylinder, the radius of the sphere (25)

We have to have  (((h/2))^2) + (r^2) = (R^2) = (25^2) for the connection between the radii of the sphere and cylinder.

Clear[r, h, v];
v = π (r^2) h;
r = (Sqrt[(25^2) - (((h/2))^2)]);
v
D[v, h]

Solve[D[v, h] == 0]

The math problem is: Find the Max of  v = h (625 - ((h^2)/4)) π  for  0 ≤ h ≤ 50.  The volume at the endpoints is zero and there is only one interior critical point:

v /. h -> 0

v /. h -> 50

v /. h -> 50/(Sqrt[3])

or solve this way:

Clear[v, h];
v[h_] = h (625 - ((h^2)/4)) π;
v'[h]

Solve[v'[h] == 0]

v[0]
v[50]
v[50/Sqrt[3]]

6 Curve Sketching

Sketch the solutions of the differential equation  (dy/dt) = y (3 - y)  that begin at y[0]=0, y[0]=1, y[0]=2, y[0]=3, y[0]=4 shown by the dots below all on the same graph.

curve = 
      Plot[0, {t, 0, 10}, AxesLabel -> {"x", "y"}, 
        PlotRange -> {-1, 5}]
Show[curve, 
        Graphics[{AbsolutePointSize[4], 
            Map[Point[#] &, {{0, 0}, {0, 1}, {0, 2}, {0, 3}, {0, 
                  4}}]}]];

Clear[y, t, soln, InitialValues];
InitialValues = {0, 1, 2, 3, 4};
soln = 
      Map[NDSolve[{y[0] == #, y'[t] == 
                y[t] (3 - y[t])}, {y[t]}, {t, 0, 10}] &, 
        InitialValues]

Plot[Evaluate[y[t] /. soln], {t, 0, 10}, 
        PlotRange -> {-1, 5}, 
        PlotStyle -> {Hue[ .1], Hue[ .2], Hue[ .3], Hue[ .4], 
            Hue[ .5]}];

7 Curve Sketching

Read the whole problem before you start work; parts (a), (b), (c) should fit together in yor answer.  
A function  f[x]  is twice differentiable for all  x.

a)  The first derivative satisfies:  f'[x] = 0 only for x = -2  and  x = +3  and nowhere else.  In addition,  f'[-4]  < 0, f'[0] > 0, and  f'[4] > 0.  Make a slope table for this information.

b)  The second derivative satifies:  f''[x] = 0  for x = +1 and  x = +3  and nowhere else.  In addition, f''[-2] > 0, f''[2] < 0, f''[5] > 0.  Make a bending table for this information.

c)  The function has values:  f[-2] = -2, f[+1] = -1, f[+3] = +1.  Make a position table for these values.

d) Use the information of parts (a), (b), (c) to plot this function as accurately as you can.

8Differentiation Rules

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

9 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.

10 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.)

Clear[f, c, x];
f[x_] = (((1 - (x^2)))^2) + 2 (c^2) (x^2)
Simplify[f'[x]]
Solve[f'[x] == 0, x]

Notice that if -1<c<1,  ((f'[x] = 4 x( x)^2)-(1 - (c^2))), so if  (x^2) > 1 - (c^2), the second term ((( x)^2)-(1 - (c^2))) is positive, while if  (x^2) < 1 - (c^2), the second term is negative, this makes the signs of  f'[x]  go negative, positive, negative, positive and the graph  y = f[x]  go down, over, up, over, down, over, up.  (Make a slope table.)

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])).

The candidates for max and min are the critical values and the endpoints.  
	f[±1] = 2  (c^2)

\(TraditionalForm\`f[\(±\@\(1 - c\^2\)\)] = \(\((1 - \((1 - c\^2)\))\)\^2 + 
            2 \(\( c\^2\)(1 - c\^2)\) = \(2  c\^2 - c\^4 = \(c\^2\)(
              2 - c\^2)\)\)\)

f[0] = 1

The question is: Which of these is largest and smallest?  If  c < (1/(Sqrt[2])), then  2  (c^2) < 1 and the maximum occurs at  x=0.

If  c > (1/(Sqrt[2])), then  2  (c^2) > 1and the maximum occurs at the endpoints.

When is  2  (c^2) < (c^2)(2 - (c^2))?  We know from the shape table down over up over down over up, that this can never happen, the interior critical points are "local" minima.  You can prove it algebraically:

2  (c^2) < (c^2)(2 - (c^2))

2 < (2 - (c^2))  (assuming c≠0, as given)

0 < -(c^2)   which never happens.

c = 1/2;
Plot[f[x], {x, -1, 1}];

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.

c = 10/11;
Plot[f[x], {x, -1, 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.

In this case there is only one real critical point  x=0.  The slope table is down over up, so the endpoints are max and the interior cirtical point is min.

c = 11/10;
Plot[f[x], {x, -1, 1}];

11  Shapes and Calculus

The graph of  y=x (x - 6) (((x - 2))^2)= -24 x+28 (x^2)-10 (x^3)+(x^4)  is shown below.

The x-values indicated at the top are the ONLY places where  (dy/dx)= 0  (x ≈ 0.61, 2.0, 4.9).

The x-values indicated at the bottom are the ONLY places where  (((d^2) y)/(dx^2)) =  0  (x ≈ 1.2, 3.8).

Make a complete shape table for this function showing where y = f[x]  is increasing and decreasing and where  y = f[x]  is convex up or down.

Clear[x, y];
y[x_] = x (x - 6) (((x - 2))^2)
Expand[y[x]]
Solve[(∂_x) y[x] == 0]
\(Solve[∂\_{x, 2}y[x] == 0, x]\)
NSolve[(∂_x) y[x] == 0]
\(NSolve[∂\_{x, 2}y[x] == 0, x]\)
curve = 
        Plot[y[x], {x, -1, 6.5}, 
          AxesLabel -> {"x", "y"}];

Show[curve, 
        Graphics[{Line[{{ .614, y[ .614]}, { .614, 55}}], 
            Line[{{2. , y[2. ]}, {2. , 55}}], 
            Line[{{4.89, y[4.89]}, {4.89, 55}}], 
            Line[{{1.24, y[1.24]}, {1.24, -55}}], 
            Line[{{3.76, y[3.76]}, {3.76, -55}}]}], 
        Graphics[{Text[".61", { .614, 65}], Text[2. , {2, 65}], 
            Text[1.2, {1.24, -65}], Text[3.8, {3.76, -65}], 
            Text[4.9, {4.89, 65}]}]];

11 TOO 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}]}]];

12 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.  USE THE SHAPE and calculus to make your selection. (NOTE:  Recall that (e^x) is always positive.)

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}}]];

12 TOO

Say which of the graphs below is the best representation of the SHAPE of the graph  y = 6 (x^5) - 10 (x^3)  AND give CALCULUS reasons why the others are NOT the correct shape.

y = 6 (x^5) - 10 (x^3);
a  =  
        Plot[y, {x, -3, 3},  Ticks -> {{-1, 1}, {}}, 
          PlotLabel -> "a"];
y = 5 (x^4) - 8 (x^2);
b  =  
        Plot[y, {x, -3, 3},  Ticks -> {{-1, 1}, {}}, 
          PlotLabel -> "b"];
y = 4 (x^3) - 6 x;
c  =  
        Plot[y, {x, -3, 3}, Ticks -> {{-1, 1}, {}}, 
          PlotLabel -> "c"];
Show[GraphicsArray[{a, b, c}]];