Let f be the function given by f(x) = 2x(e^2x)?

Question

(a) Find limit as x approaches negative infiniti for f(x) and limit as x approaches infiniti for f(x)

(b) Find the Absolute minimum value of f.

(c) What is the range of f?

(d) Consider the family of functions defined by y = bx(e^bx), where b is a nonzero constant. Show that the absolute minimum value of bx(e^bx) is the same for all nonzero values of b.

Answer 1

(a) I assume you mean

f(x) = 2x e^(2x)

Let's take the limit as x approaches -infinity. Note the behavior of e^x; as it gets really large, e^x gets really large, and as x gets large negatively, e^x approaches 0. Therefore, the limit of e^x as x approaches negative infinity is equal to 0. e^(2x) works similarly.

Now, let's solve for the limit.

lim [2x * e^(2x)]
x -> -infinity

First thing to note is that you get the form [infinity*0], which is an indeterminate form. Since that's the case, you have to make this into a fraction to use L'Hospital's rule, and one way to do that is to bring the e^(2x) down to the denominator. Note that anything with a power that gets brought down to the denominator has the power negated.

lim [2x / e^(-2x)]
x -> -infinity

This is of the form [infinity/infinity], so we use L'Hospital's rule. Take the derivative of top and bottom to form a new limit.

lim [2 / (-2 e^(-2x))]
x -> -infinity

Let's pull out the constants.

(-2/2) * lim [1 / e^(-2x)]
x -> -infinity

And now, let's simplify, and make the power of e positive again.

(-1) * lim [e^(2x)]
x -> -infinity

As per the behaviour of e^(2x), the limit of that is equal to 0. So we now have

(-1) (0) = 0

The limit as x approaches infinity makes the function get really large, so the limit doesn't exist.

(b)

To solve for the absolute minimum (if it exists), we find the local extreme by finding the first derivative f'(x) and equating f'(x) to 0.

f(x) = 2x (e^(2x)). Using the product rule and chain rule,

f'(x) = [2][e^(2x)] + [2x] [2e^(2x)]. Factoring out 2e^(2x), we have
f'(x) = [2e^(2x)] [1 + 2x]. Now, making f'(x) = 0,

0 = [2e^(2x)] [1 + 2x]

Which means we equate both of those bracketed terms to 0. There will be no solution for 2e^(2x) = 0 {I'll leave that part as homework for you. We can equate the other one to 0, though.

1 + 2x = 0, means
2x = -1, or
x = -1/2

Now, we determine if it is a local minimum or maximum by testing a value greater than -1/2 and less than -1/2 for
f'(x) = [2e^(2x)] [1 + 2x]. What we're testing for is positivity/negativity.

Test a value greater than -1/2, 0:
f'(x) = [positive][positive] = positive. Therefore, f is increasing from [-1/2, infinity).
Test a value less than -1/2, -1:
f'(x) = [positive] [negative] = negative. Therefor, f is decreasing on (-infinity, -1/2]

That means we have a local minimum at x = -1/2.
The actual local minimum would be
f(-1/2) = 2(-1/2)e^(2*(-1/2)) = (-1)e^(-1) = -1/e

f is decreasing from (-infinity, -1/2]
f is increasing from [1/2, infinity)

Since it is the ONLY minimum, it is the absolute minimum.

(c) The range of f is dependent on the absolute minimum value. Since we know what the minimum is, f(x) would be defined from that minimum point onward. The range is [-1/e, infinity).

I'll let you solve (d) on your own.

Answer 2

a) for the limits at infinity, you can rearrange the function into a ratio and apply L'Hopital's rule. Though it should be obvious from inspection that the limit at positive infinity does not exist, you need to use L'Hopital's rule to show it if you mean to do the problem correctly. Doing the limit at negative infinity in my head, I am getting 0.

b) this will be the value of the limit at negative infinity.

c) the range starts with the answer to (b) and goes up to infinity. Don't include the lower endpoint, because that value is a limit which the function does not actually acheive for any real value of the variable.

d) this is a simple application of L'Hopital's rule, where you do the same thing as in part (a) but with "b" in place of "2".

Answer 3

hi, permit f be the function given via f(x) = |x|. Which of right here statements approximately f are authentic? a million. f is non-give up at x = 0 2. f is differential at x = 0 3. f has an absolute minimum at x = 0 <==answer The graph is V formed, beginning up from its vertex at (0,0). for this reason, 0 is an absolute minimum of the graph. i desire that enables!! :-)

Answer 4

\d\ y=bx*exp(bx);
formula: if y=uv, then y’ = uv’ +u’v;
formula: if y=f(u), then ‘y = (df/du)*u’
y’ = bx*exp(bx)*b + b*exp(bx) = b* exp(bx) * (bx+1) = 0; b*exp(bx) is never =0, thus x1=-1/b and y(x1) = b*(-1/b) *exp(-b/b) = -1/e, that is does not depend on b;
Now find y’’; here xy’ = y*(bx+1); derive both parts! y’+xy’’ = yb +y’bx + y’ or y’’ = b(y+y’)/x and y’’(x1) = (-b/e +0)/(-1/b) = b^2/e > 0 means local min!
Now if bx - - > -inf, then y - - > 0, and 0 is greater than –1/e
Now if bx - - > +inf, then y - - > +inf; that means the point [-1/b, -1/e] is also the abs min!