How do I solve this step by step?

Question

For the function 𝑔(𝑥) = 𝑥𝑒^-2x find all critical points and say which are local maximums or minimums.

Also fine the point of inflection and say where the function is concave up and where it is concave down.

Answer 1

y = x exp(-2x)

y' = -2x * exp(-2x) + exp(-2x) = exp(-2x) [ -2x + 1] = exp(-2x) [ 1 - 2x]

The derivative is zero when 1-2x = 0 because the exponential is never zero

x = 1/2 is the critical point ---> f(1/2) = 1/2 * exp( -2 * 1/2) = 1/2 * exp (-1) = 1/(2e)

the critical point is ( 1/2, 1/(2e) )

y'' = (-2) exp(-2x) + (-2)exp(-2x)[1 - 2x]

= (-2) exp(-2x) [ 1 + 1 - 2x]

= (-2) exp(-2x) [2-2x]

again the exponential is never zero, so the flex value is 2-2x = 0 ---> x=1

the concavity point is (1, 1/e^2)

Answer 2

The critical points are found by finding the x-values for which the derivative equals zero or for which the derivative does not exist.

This function is in the form of u•v where u = x and v = e^-2x. (two functions multiplied together) and so to find the derivative, we must use the product rule that says where (in this case) g(x) = u•v then g'(x) = u'v + uv'

So, since u = x, then u' = 1 and since v = e^-2x, v' = -2e^-2x (by the chain rule)

So:

g'(x) = u'v + uv'

g'(x) = (1)(e^-2x) + (x)(-2e^-2x)

g'(x) = e^-2x - 2xe^-2x

g'(x) = e^-2x(1 - 2x)

This function exists for all values of x, so we just need to find where it is equal to zero. Using the zero product rule, for g'(x) to be equal to zero, then either e^-2x = 0 or 1 - 2x = 0. e^-2x can never be zero, so we only need to find when 1 - 2x = 0

1 - 2x = 0

1 = 2x

x = 1/2

So the only critical point is x = 1/2

Since we are also looking for the points of inflection, we will need to take the second derivative so I'll wait to determine if x = 1/2 is a max or min by looking at the second derivative.

To find g'', since g'(x) = e^-2x(1 - 2x) we also will need to use the product rule, this time u = e^-2x and v = 1 - 2x

So u' = -2e^-2x and v' = -2

g''(x) = u'v + uv'

g''(x) = (-2e^-2x)(1 - 2x) + (e^-2x)(-2)

g''(x) = -2e^-2x + 4xe^-2x - 2e^-2x

g''(x) = -4e^-2x + 4xe^-2x

g''(x) = 4e^-2x(x - 1)

Again, setting this to zero gives us only x - 1 = 0 or x = 1 as a POSSIBLE point of inflection. To check, let's perform the 2nd derivative test.

We draw a number line and put the critical point and the possible point of inflection on it:

And then we evaluate the sign of g'' in the various intervals shown, like this:

So, x = 0.5 must be a relative maximum (since it is concave down on the interval surrounding it. And there is a point of inflection at x = 1 because it changes concavity there.

Answer 3

You will have to find the 1st and 2nd derivatives and related critical and "hyper" critical numbers. Actually I do not think textbooks use the term "hyper"critical anymore, but those are numbers where 2nd derivative is zero.

g'(x) = x (e^-2x) (-2) + e^-2x (found by using product rule.

To see what is going on it helps to factor out the e^-2x.

= e^-2x(-2x + 1)

Since e^-2x is never equal to zero the only critical number happens when -2x+1` = 0, or when x = 1/2.

Since g'(0) = 1 which is positive and g'(1) = a negative number (confirm), the function goes from increasing to decreasing at x = 1/2. That means (1/2, g(1/2)) is a relative max.

g''(x) = -2(e^-2x) + (-2x+1)(e^-2x)(-2) (again the product rule on the highlighted version of g' shown above.

= -2e^-2x (1 + (-2x) + 1)

= -2e^-2x (-2x +2)

Once again this only equal zero when -2x+2 = 0 -> x= 1

I will let you confirm that there is a point of inflection at (1, g(1)) by testing the 2nd derivative to either side of x=1.

You will discover that the indeed there is a point of inflection at (1,g(1)). .

Visit this URL for confirmation.

desmos.com/calculator/qb4yqgpjgd