Hello how do I plot plot function exp^(-x), x <= 0 and exp^(x), x >0, Thanks alot!

Hello how do I plot plot function exp^(-x), x <= 0 and exp^(x), x >0, Thanks alot!

This is what I have:

 x = 0:0.1:10; y1=exp(-x); y2=exp(x); plot(x,y1,x,y2)

[Walter Roberson]

hint: exp(abs(x))

Answer 1

[KALYAN ACHARJYA]

x=-10:0.1:10;
y1=exp(-x);
plot(x,y1);
hold on;
y2=exp(x);
plot(x,y2);

[Walter Roberson]

This is not correct. Consider x = -2, then according to the problem definition since x < 0, y = exp(-x) which is exp(-(-2)) which is exp(2) -- and only exp(2), no exp(-2) involved. Each x has exactly one y, not two y values.

Answer 2

[Kristy Kwok]

Thanks Walter!

This is what I have:

x = 0:0.1:10;
y1=exp(-x);
y2=exp(x);
plot(x,y1,x,y2)

[Walter Roberson]

No, this is no right.

1. You do not plot with negative x.
2. the equations from the question are set up such that exp() of a negative number is never taken, but you take exp(-x) for positive x.

Answer 3

[Kristy Kwok]

Thanks Walter!

I have come up with this:

x = 0:0.1:10;
y1=exp(abs(x));
y2=exp(x);
plot(x,y1,x,y2)

But I wasnt sure what to do with the x<=0 and X >0

[Walter Roberson]

Your plot has no negative x values.

If you use abs(x) you do not need two different functions.

Answer 4

[Kristy Kwok]

Hi Walter,

I understand exp (abs(x)) is the same as exp(x) but I wasnt sure how to specify the limits x<=0 and X >0?

Would that be is the range of X? So it would be like this:

x = -3 :0.1:10;
y=exp(abs(x));
plot(x,y)

?

[Walter Roberson]

Yes, like that.

The symmetry of it is not apparent because you are going only to 3 to the left of x = 0 but to 10 to the right of x = 0.

[Kristy Kwok]

Thank you so much Walter!! Have a great day!:D

I have the code like this: x=-10:0.1:10; y1=exp(-x); plot(x,y1);

Answer 5

[Kristy Kwok]

Hello Kalyan,

Thank you so much! I am sorry but I am still lost on how to specify the limits x<=0 and X >0, can you please briefly explain?

Many thanks!

Answer 6

[Walter Roberson]

Sigh. Once more with making the different ranges explicit:

x = -3 :0.1:10;

y1 = zeros(size(x));

for K = 1 : length(x)
   this_x = x(K);
   if this_x < 0
     y1(K) = exp(-this_x);
   else
    y1(K) = exp(this_x);
   end
end

%Which can also be written as

y2 = zeros(size(x));
mask = x < 0;
y2(mask) = exp(-x(mask));
y2(~mask) = exp(x(~mask));

%and can also be written

x = -3 :0.1:10;
y3 = exp(abs(x));

subplot(1,4,1)
plot(x, y1);
title('loop')

subplot(1,4,2)
plot(x, y2)
title('logical indexing');

subplot(1,4,3)
plot(x, y3)
title('abs()')

subplot(1,4,4)
plot(x, y1-y2, 'b*-', x, y1-y3, 'g.-')
title('difference between methods. Yes, it is EXACTLY 0')
legend('loop minus mask', 'loop minus abs')

fprintf('maximum difference between loop and mask: %g\n', max(abs(y1-y2)));
fprintf('maximum difference between loop and abs: %g\n', max(abs(y1-y3)));
fprintf('maximum difference between mask and abs: %g\n', max(abs(y2-y3)));