1. Help with Derivatives.

I have a mid-term tomorrow and still don't understand this specific concept.

Let f(x)=(x^4-18x^2)/5

(a) Use the definition of a derivative or the derivative rules to find f'(x).

(b) Use the definition of a derivative or the derivative rules to find f' '(x)=

My answer: f' '(x) = (12x^2-36)/5

(c) On what interval is f increasing (include the endpoints in the interval)?
interval of increasing =

(d) On what interval is f decreasing (include the endpoints in the interval)?
interval of increasing =

(e) On what interval is f concave downward (include the endpoints in the interval)?
interval of increasing =

(f) On what interval is f concave upward (include the endpoints in the interval)?
interval of increasing =

2. Re: Help with Derivatives.

I have figured out (c) and (d) for you. Maybe someone else that knows more can help you with (e) and (f).

f' = (4x^3 - 36x)/5

To figure out where its decreasing or increasing, we need to find the critical point. Which is found by setting the first dervative equal to 0, and then solving for x.

0 = (4x^3 - 36x)/5

0 = x(4x^2 - 36)
0 = x(2x+6)(2x-6)

So x = 0, x = -3, x = 3 are our options.
To classify them as a critical point we need to test points within our critical points, and observe the behavior when we plug them into f'(x)... observing meaning just check whether they are + or -
f'(-4) = -
f'(-1) = +
f'(1) = -
f'(4) = +

So we are decreasing from x = -∞ to x= -3
Then increasing from x = -3 to x = 0
Then decreasing from x= 0 to x = 3
Then increasing from x = 3 to x = ∞

So our interval of decreasing:
(-∞, -3] U [0, 3]
And our interval of increasing:
[-3, 0] U [3, ∞)

3. Re: Help with Derivatives.

The function f is increasing if $f'(x)>0$, decreasing if $f'(x)<0$, concave upward if $f''(x)>0$, and concave downward if $f''(x)<0$. Since you've calculated the first and second derivatives already, it's only a matter of determining where they're positive and negative.

- Hollywood

4. Re: Help with Derivatives.

Let f(t) = 9t^4 - 7t^2 + 12t -4 , Find (laplace) L[d^2f/dt^2]

I get answer 108(2/s^3) - 14/s