Find the inflection points

I have given the function and it asks me to indicate the points of inflection and where the function is concave up(CU) or concave down(CD);

I proceded with the first derivative f'(x)=e^(13x)*x(13x+2),

- the second f"(x)=e^(13x)*[169x^(2)+52x+2],

since e^x is never 0, this one gives me[169x^(2)+52x+2]=0 for x=[-(52)+(sqrt(1352))]/(2*169) and x=[-(52)-(sqrt(1352))]/(2*169);

it bugs that those points are quite bizarre and on all the intervals the function gives me CU, does that mean that I don't have inflection points here, I might have mistaken something...

Thanks

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Re: Find the inflection points

Re: Find the inflection points

Just one question -

do we have to evaluate the second derivative on the first derivative's critical points; I have seen the examples where we find the critical points for the second derivative and evaluate it on them.... I'm confused...

Re: Find the inflection points

The steps are as under:

Step 1. find the derivative f'(x)

Step-2. Equate f'(x) = 0 and solve. These are the values of x where the function may have a maxima ( CD) or Minima ( CU)

Step 3. Find f"(x)

Step 4. Find the value of f"(x) for the values of x found in step 2.

step 5. The values of x for which f"(x) is < 0 the function will have a maxima ( CD) and the values of x for which f"(x) > 0 the function will have a minima ( CU).

Step 6. If one is to find the maximum or minimum value then we have to evaluate f(x) at those values of x [ Step - 2 ]

Re: Find the inflection points

Quote:

Originally Posted by

**ibdutt** The steps are as under:

Step 1. find the derivative f'(x)

Step-2. Equate f'(x) = 0 and solve. These are the values of x where the function may have a maxima ( CD) or Minima ( CU)

Step 3. Find f"(x)

Step 4. Find the value of f"(x) for the values of x found in step 2.

step 5. The values of x for which f"(x) is < 0 the function will have a maxima ( CD) and the values of x for which f"(x) > 0 the function will have a minima ( CU).

Step 6. If one is to find the maximum or minimum value then we have to evaluate f(x) at those values of x [ Step - 2 ]

sorry, but it still gives me the wrong answer - it doesn't indicate which one is correct but it's something wrong with those points...

any help please

Re: Find the inflection points

Quote:

Originally Posted by

**dokrbb** Just one question -

.do we have to evaluate the second derivative on the first derivative's critical points; I have seen the examples where we find the critical points for the second derivative and evaluate it on them.... I'm confused...

To do **what**? What you **have** to do or **should** do depends upon what problem you are trying to solve! **If** you are trying to find max or min values then checking the second derivative at the points where the derivative is 0 will tell you whether they is a max or min there. To find "points of inflection", you want to find places where the **second** derivative changes sign- and to do that you will look for points where the second derivative is 0

Re: Find the inflection points

Quote:

Originally Posted by

**HallsofIvy** To do **what**? What you **have** to do or **should** do depends upon what problem you are trying to solve! **If** you are trying to find max or min values then checking the second derivative at the points where the derivative is 0 will tell you whether they is a max or min there. To find "points of inflection", you want to find places where the **second** derivative changes sign- and to do that you will look for points where the second derivative is 0

That's right, and that's what I tried to explain that, but in previous posts I was suggested to look at where the second derivative is "0" by evaluating it at the critical points for first derivative (non-sense in my opinion),

but well, I have to find the 2 points(A and B) of inflection and whether the function is CU or CD on the intervals (-Inf, A), (A, B), (B, Inf) ,

and I found the critical points for the second derivative: x=-0.045060495(B) and x=-0.262631812 (A),

finally, the function would be CU -> CD -> CU

The answers are correct as I found them, thanks for help... though the previous advises weren't quite precise :)

hallsofivy, I thought I was pretty clear in my first post, but I'm especially thankful for your critical posts, I like them!