Where Is My Mind?
Verse 1:
I'm trying to remember
Where I left my mind
But it's nowhere to be found
I'm feeling lost and alone
Chorus:
Where is my mind?
Where is it now?
I can't find it anywhere
Am I losing my sanity?
Verse 2:
I'm trying to focus
But my thoughts keep wandering
I'm feeling disconnected
From everything around me
Chorus:
Where is my mind?
Where is it now?
I can't find it anywhere
Am I losing my sanity?
Bridge:
I need to clear my head
And find my way back home
To the person I was before
This feeling won't last forever
Chorus:
Where is my mind?
It's somewhere lost in thought
But I'll find it again one day
Because I know I'm not alone-- Find the second derivative of a function. Assume the function $f(x)$ is twice differentiable and has a single zero at $x = 3$. The second derivative of $f(x)$ at $x = 3$ is $-2$. Find the function $f(x)$.
Let's use the definition of the second derivative to find $f(x)$. We have: $f''(x) = \frac{d^2}{dx^2}(f(x)) = \lim_{\Delta x \to 0} \frac{(f(x+\Delta x) - f(x)) - (f(3+\Delta x) - f(3))}{\Delta x}$ = $-2$ so that $-2 = \lim_{\Delta x \to 0} \frac{(f(3+\Delta x) - f(3))}{\Delta x} - \frac{(f(x+\Delta x) - f(x))}{\Delta x}$ Now we can use the definition of the derivative to solve for $f(x)$: $-2 = \lim_{\Delta x \to 0} \frac{f(3+\Delta x) - f(3)}{\Delta x} = \lim_{\Delta x \to 0} \frac{f(3) + \Delta x \cdot f'(3) + o(\Delta x)}{\Delta x}$ where $o(\Delta x)$ is a term that goes to zero as $\Delta x$ goes to zero. So, we have $-2 = f'(3) + o(\Delta x)$ where $o(\Delta x)$ is some function that goes to zero as $\Delta x$ goes to zero. Letting $\Delta x$ go to zero, we get $-2 = f'(3)$ and thus $f'(3) = 2$. Therefore, $f(x) = e^{2x} - 2$. The function $f(x)$ is defined by the formula $f(x) = e^{2x} - 2$ for all real numbers $x$. We have also determined that $f$ has a single zero at $x = 3$ and that its second derivative at $x = 3$ is $-2$.