2 Questions on Concavity and Absolute Minimum

[blitzen]

Hey all,

I am doing some review problems for an upcoming exam and I have run into two problems that I feel should be easy but I am just missing something:

1) Find the intervals on which the graph of f is concave upward/downward, and the inflection points.
------Now, to find the inflection points, I need to get to the second derivative.
f(x) = 8e^x - e^2x
f''(x) = 8e^x - 4e^2x = 4(2e^x - e^2x)

If that is correct, which I think it is, then I should solve for when x is equal to zero...I am having trouble doing that

2) Find the absolute minimum on the interval (0, infinity)
f(x) = 2x + (5/x) + (4/x^3)
------for this one I need to get the f'(x)
f'(x) = 2 + 5*ln(x) - (2/x^2) which I tried to simplify to 1 = x^2[(5/2)ln(x) + 1]

again, I do not know if that is correct....even if it is, that is where I am stuck.

Thanks for any help!

 

[Deleted member 4993]

Hey all,

I am doing some review problems for an upcoming exam and I have run into two problems that I feel should be easy but I am just missing something:

1) Find the intervals on which the graph of f is concave upward/downward, and the inflection points.
------Now, to find the inflection points, I need to get to the second derivative.
f(x) = 8e^x - e^2x
f''(x) = 8e^x - 4e^2x = 4(2e^x - e^2x)

You need to use correct grouping symbols:

f"(x) = 8e^x -4e^(2x) = 4e^x (2-e^x)


for f"(x) = 0 we have

e^x = 0

or

2 - e^x = 0

Now finish it.....

If that is correct, which I think it is, then I should solve for when x is equal to zero...I am having trouble doing that

2) Find the absolute minimum on the interval (0, infinity)
f(x) = 2x + (5/x) + (4/x^3)
------for this one I need to get the f'(x)
f'(x) = 2 + 5*ln(x) - (2/x^2) which I tried to simplify to 1 = x^2[(5/2)ln(x) + 1]

You are not differentiating - you are integrating!!!

again, I do not know if that is correct....even if it is, that is where I am stuck.

Thanks for any help!

.

 

[blitzen]

.

I am confused on how you get 8e^(x) - 4e^(2x) = 4e^(x)*(1-e^x)?

Also..on the second problem I am getting the sq. rt. of (5/2) as my answer, but according to the solutions manual that is incorrect.
f'(x) = 2 - (5/(x^2)) - (12/(x^4)) = 0
solved this down to x^(2) * [2x^(2) - 5] = 0 ??

 

[mmm4444bot]

I am confused on how you get 8e^(x) - 4e^(2x) = 4e^(x)*(1-e^x)?

Subhotosh meant to type 2 - e^x (fixed), and here's a hint regarding the factorization (the hint is one of the properties of exponents):

b^(m*n) = (b^m)^n = (b^n)^m

 

[mmm4444bot]

2 - 5/(x^2) - 12/(x^4) = 0

[simplified] this down to x^2 * [2x^2 - 5] = 0

This factorization is not correct.

I would start by using a common denominator to combine 2 - 5/(x^2) - 12/(x^4) into a single ratio, but that's me. :cool:

The numerator of that ratio should have quadratic form, I'm thinking.

 

[blitzen]

So if I am not mistaken I should make that (2x^(4) - 5x^(2) - 12) / x^(4)

I still need to make this equal to zero in order to find any critical numbers.
That being said I simplified it to [x^(2)]*[x^(2) - (5/2)] = 6
Further to [2x^(4)] - 5x^(2) - 12 = 0
From there I did the quadratic formula to obtain values 4 and -(3/2). I substituted in those values to the original f(x) and results indicated an absolute minimum at x = (-3/2) ...roughly -7.52

Once I looked at the solutions manual I was again incorrect..

 

[Deleted member 4993]

Another way would be to substitute:

u = x^-2

then we have:

12u² + 5u -2 = 0

(4u - 1)(3u + 2) = 0

Now finish it......

 

[Deleted member 4993]

So if I am not mistaken I should make that (2x^(4) - 5x^(2) - 12) / x^(4)

I still need to make this equal to zero in order to find any critical numbers.
That being said I simplified it to [x^(2)]*[x^(2) - (5/2)] = 6
Further to [2x^(4)] - 5x^(2) - 12 = 0
From there I did the quadratic formula to obtain values 4 and -(3/2).

Those are values of x² - you need values of x

I substituted in those values to the original f(x) and results indicated an absolute minimum at x = (-3/2) ...roughly -7.52

Once I looked at the solutions manual I was again incorrect..

.