# Thread: second derivative/point of inflection

1. ## second derivative/point of inflection

Given f(x) = e^(-5x) sin(1x) , 0< x < pi/2

a) Find the second derivative

b) suppose that f has a point of inflection at x=c. Using the result of a), you can formulate the equation for c in the form:
tan 1c = K
where K is some constant.
Find the constant K as well as the points of inflection

C) Find the (open) intervals of concavity. If there is more than one interval enter using the union symbol U. Enter the empty set as {}
"f is concave upward on open intervals _________"
"f is concave downward on open intervals _________"

Alright so i found the second derivative:
f''(x) = e^(-5x) * (24sinx - 10cosx)

and in order for there to be inflection points where x = c, f''(c) = 0
I also know that for finding concavity(part c) i need to test points, which i can do on my own once i figure out part b!
I'm stuck on isolating "c" , i know this seems like grade 2 stuff by now, but still.
here's where i'm at ..

0 = e^(-5x) * (24sinx - 10cosx)
0 = ln[e^(-5x)] * (24sinx - 10cosx)
0 = (-5x) ln(e) * (24sinx - 10cosx)

we replace x with c..
0 = (-5c) ln(e) * (24sinc- 10cosc)

but i get so lost here. How do i isolate??! obviously i'm missing something/have done something wrong.

Britt

2. Originally Posted by williamb
Given f(x) = e^(-5x) sin(1x) , 0< x < pi/2

a) Find the second derivative

b) suppose that f has a point of inflection at x=c. Using the result of a), you can formulate the equation for c in the form:
tan 1c = K
where K is some constant.
Find the constant K as well as the points of inflection

C) Find the (open) intervals of concavity. If there is more than one interval enter using the union symbol U. Enter the empty set as {}
"f is concave upward on open intervals _________"
"f is concave downward on open intervals _________"

Alright so i found the second derivative:
f''(x) = e^(-5x) * (24sinx - 10cosx)

and in order for there to be inflection points where x = c, f''(c) = 0
I also know that for finding concavity(part c) i need to test points, which i can do on my own once i figure out part b!
I'm stuck on isolating "c" , i know this seems like grade 2 stuff by now, but still.
here's where i'm at ..

0 = e^(-5x) * (24sinx - 10cosx)
0 = ln[e^(-5x)] * (24sinx - 10cosx)
0 = (-5x) ln(e) * (24sinx - 10cosx)

we replace x with c..
0 = (-5c) ln(e) * (24sinc- 10cosc)

but i get so lost here. How do i isolate??! obviously i'm missing something/have done something wrong.

Britt
Okay, so your second derivative is correct, and you want to know when it is eqaul to zero so...

$\displaystyle 0=e^{-5x}[24\sin(x)-10\cos(x)]$

so from here use the zero factor rule

so we get

$\displaystyle e^{-5x}=0$ or $\displaystyle 24\sin(x)-10\cos(x)=0$

Since e never equals zero the first has no solutions but

$\displaystyle 24\sin(x)-10\cos(x)=0 \iff 24\sin(x)=10\cos(x) \iff$
$\displaystyle \tan(x)=\frac{10}{24} \iff x=\tan^{-1}\left( \frac{5}{12}\right)$

3. AH BUT OF COURSE!
It's been a long day,
you're greatly appreciated thankyou!

4. AH dang i spoke too soon
So to find concavity, i need to find the first and second derivatives and equate to zero in order to find the critical and inflection points right?

well if f'(x) = e^-5x (-5sinx + cosx)
then if we set to zero, how do we find these points?
(-5sinx + cosx) = 0
sinx = 0 at pi, cosx = 0 at pi/2 and 3pi/2
do i just use number values of those?