Homework # 5

5.1

(a) Graphically, I approximated the values of the functions at zero to  -1.63 and 5.63

Matlab code:

x  = (-2:0.01:6);

f_x = @(x) -0.6*x.^2 + 2.4*x + 5.5;

plot (x,f_x(x))

(b) Using the quadratic formula:

I was able to obtain the values at :  -1.6286 and 5.6286

Matlab:

a = -0.6;

b = 2.4;

c = 5.5;

quad1 = ((-b + sqrt(b^2 -4*a*c))/(2*a))

quad2 = ((-b - sqrt(b^2 -4*a*c))/(2*a))

(c) Using the bisecting method I got the solution of 5.62
Matlab:

f_x = @(x) -.06*x.^2 +2.4.*x +5.5;

%x lower

x_l = 5

%x upper

x_u = 10

x_r = (x_l * x_u) / 2

for i = 1:3

    if ((f_x(x_l))*(f_x(x_u)) < 0)

        x_u = x_r;

    elseif ((f_x(x_l))*(f_x(x_u)) > 0)

        x_l = x_r;

    elseif ((f_x(x_l))*(f_x(x_u)) == 0) 

       

        break;

    

    end

    x_r =(x_l+x_u)/2;

end

5.6

(a) This problem is the same as 5.1. Here is the graphical solution with my approximations at :  1

Matlab:

x  = (-2:0.001:2);

f_x = @(x) log(x.^4);

plot (x,f_x(x))

grid on

(b) This is also the same as problem 5.1 and the approximation is = 0.5 

Matlab: 

x  = (-2:0.001:2);

f_x = @(x) log(x.^4);

%x lower

x_l = .5

%x upper

x_u = 2

x_r = (x_l * x_u) / 2

for i = 1:3

    if ((f_x(x_l))*(f_x(x_u)) < 0)

        x_u = x_r;

    elseif ((f_x(x_l))*(f_x(x_u)) > 0)

        x_l = x_r;

    elseif ((f_x(x_l))*(f_x(x_u)) == 0) 

       

        break;

    

    end

    x_r =(x_l+x_u)/2;

end

(c)For this part I used the Method of False Position:

I coded it up in Matlab for three iterations and got these results:

Output:

1st Iteration: 1.2500

2nd Iteration: 1.0674

3rd Iteration:  1.0186  

Matlab:

x  = (-2:0.001:2);

f_x = @(x) log(x.^4);

%plot (x,f_x(x),'red')

%grid on

x_u = .5;

x_l = 2;

for iter = 1:3

x_r = x_u - (f_x(x_u)*(x_l - x_u))/ (f_x(x_l) - f_x(x_u));

x_l = x_r ;

x_r

end

5.7

(a)To find the root analytically I used Wolfram alpha and got the value 2.6666.....

(b)Graphically I approximated the value to 2.6

Matlab:

x  = (-2:0.1:6);

f_x = @(x) (0.8 - 0.3.*x)./x;

plot (x,f_x(x),'red')

grid on

(c)For this part I used the Method of False Position:

I coded it up in Matlab for three iterations and got these results:

Output:

1st Iteration: 2.8750 

True error: -0.0781

Approximate error: 0.0435 

2nd Iteration: 2.7969

True error: -0.0488

Approximate error: 0.0726 

3rd Iteration:  2.7480

True error: -0.0305

Approximate error: 0.0917

Matlab:

x  = (-2:0.1:6);

f_x = @(x) (0.8 - 0.3.*x)./x;

x_u = 1;

x_l = 3;

x_rp = x_l;

true = (8/3);

for iter = 1:3

x_r = x_u - (f_x(x_u)*(x_l - x_u))/ (f_x(x_l) - f_x(x_u));

x_l = x_r ;

%true error 

err_t(iter) = (((true - x_r)/true)*1); 

%approx error

err_a(iter) = abs((((x_r - x_rp)/x_r)*1));

sol(iter) = x_r;

end

sol

err_t

err_a

5.13

For this problem I used the same algorithm from the last few problems to execute the false position method. I kept using different initial values until my approximate error was close to 0.1% which gave me a value of 59.8428.

Matlab:

g = 9.8;

c = 15;

v = 35;

t = 9;

m = (25:1:70);

f_m = @(m) ((g*m)/c).* (1 - exp(-((c./m)*t)))-v;

plot (m,f_m(m));

x_u = 20;

x_l = 60;

x_rpre = x_l;

for iter = 1:5

x_r(iter) = x_u - (f_m(x_u)*(x_l - x_u))/ (f_m(x_l) - f_m(x_u));

x_l = x_r ;

%app error 

if (iter > 1)

    err_a(iter) = abs(((x_r - x_rpre)/x_r))*1;

end 

end

x_r

err_a

5.21

For this problem I simply plugged in the values given and solved the function with Matlab 

Matlab:

A = 5500;

n = 6;

P = 25000;

    

i = (-5:0.1:10);

int = @(i) P* ((i(1+i)^n)/((1+i)^n - 1));

x_u = 1;

x_l = 3;

for iter = 1:3

x_r = x_u - (int(x_u)*(x_l - x_u))/ (int(x_l) - f_x(x_u));

x_l = x_r ;

sol(iter) = x_r;

end

sol