# Thread: Find the inflection points

1. ## Find the inflection points

I have given the function $f(x)=x^{2}e^{13x}$ 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

3. ## 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...

4. ## 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 ]

5. ## Re: Find the inflection points

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...

6. ## Re: Find the inflection points

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

7. ## Re: Find the inflection points

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!