Question

One calc question

Answer 1

l

Answer 2

Qyestion 1 or 2

Answer 3

@doulikepiecauseidont both

Answer 4

for q#1 :
second derivative has to be 0 at inflection points,
so find the second derivative and set it equal to 0 :

http://www.wolframalpha.com/input/?i=%28x+%2B++log_3%28x%5E2%2B5%29%29%27%27%3D0

Answer 5

Ganeshie showed you how to do the first one, the second is comparatively simple. Just note that to find the derivative of a function, rewrite it as a quotient of natural logarithms:
\[y =log_a(b)=\frac{ln (b)}{ln(a)}\]

For #2, you're asked to find extrema of
\[f(x) = 20x^2e^{(-3x)}, x\in [0, \infty) \]So it's pretty clear the whole exercise is just in computing pain in the retricederivatives by hand. The good news is that this is pretty much as arduous and terrible as calculus can possibly be made, and you're surviving it. The bad news is that we have a function of the form
\[f(x) = u(x)*v (w(x))\] and its derivative will be of the form \[f'(x) = u'(x)*v(w(x))+u(x)*v'(w(x))*w'(x).\]To straighten it out in your head, let's explicitly device u, v and w:
\[u =20x^2\\v=e^w\\w=-3x\]and our derivatives,
\[u' =40x\\v' = e^w * w'\\w' = -3\]
So let's fill in for that ugly crap above, and we get
\[f' = 40x*e^{-3x}+(20x^2)(-3)(e^{-3x})\\\quad=e^{-3x}(40x-60x^2)\\\quad=20e^{-3x}(2x-3x^2)
\]
Now that we have a workable function, we see f has critical points at x = 0 and 2/3. The rest should be fairly easy depending on what method you want to use--based on the choices, you only need to check whether x = 0 is a minimum.

Answer 6

explicitly define* u, v, w wow I'm tired

Answer 7

@bhl6180 could u help me with a few more

Answer 8

Sure, give me a few

Answer 9

k thnxs

Answer 10

Wow... ok, the first one you sent is the "area of a rectangle" which is basically just asking you whether the indefinite integral diverges, as far as I can tell. Note that
\[\int\limits_{0}^{\infty} \frac{ln(x^2+1)}{x} dx>\int\limits_{0}^{\infty} \frac{ln(x)}{x} dx > \int\limits_{0}^{\infty} \frac1x dx
\]so it diverges. So that'd be no maximum.

Answer 11

#3 is the function I did a little while ago, just take another derivative the same way I showed you and find the zeros. The second derivative is
\[40e^{-3x}-240xe^{-3x}+180x^2e^{-3x}\]according to MATLAB, which is good enough for me. The solutions to that equation are transcendental as far as I know, so you just need to plug it into a calculator to find the zeros numerically

Answer 12

wow thankyou

Answer 13

Unless I'm missing something with #4, the one with the trout, it gives you a strict monotone increasing power function as the number of trout "still alive" after t years... those trout are really banging the bejesus out of each other. That means the population will never be 2000.

I just wasted a bunch of time trying to work this out if the function was supposed to be a negative power of 9 but it still doesn't work, so I'm sort of at a loss on this.

Answer 14

ok thankyou all help is appreciated

Answer 15

#8 is a lot more straightforward (or so it would seem... just scroll to the bottom), considering the last one made no sense and probably had one or more typos in it. We're given a position function
\[s(t) = e^{12t}\cos(5t), t> 0
\]We're told its velocity attains a relative max or min at T seconds--when you see "a relative max or min" it should set off alarm bells in your head saying "find where the derivative equals zero."

HOWEVER, this one sprinkles on a bit more bs because you're given a position function, not the velocity. So first we need to find the velocity function, which you can probably do in your head:
\[s'(t) = v(t) = 12e^{12t}\cos(5t) -5e^{12t}\sin(5t)\]
Then since we're finding the relative mins and maxes, we need the derivative of the velocity function (the acceleration)--this one we'll use MATLAB because lol arithmetic:
\[v'(t) = 119\cos(5t)e^{12t}-120\sin(5t)e^{12t}=0
\]We can do some algebra to make this look a bit nicer, and get
\[v'(t) = e^{-12t}[119\cos(5t) - 120\sin(5t)]
\]
Clearly, this equation equals zero when 119cos(5t) = 120sin(5t), so let's algebra things up some more
\[119\cos(5t) = 120\sin(5t)\\119 = 120 \frac{\sin(5t)}{\cos(5t)}\\\frac{119}{120}=\tan(5t)
\]
And now we have a problem, namely that whoever is writing this homework or exam or whatever is apparently not too bright, because once again this isn't one of the options. However, one of the options is
\[\tan(5t)=2.4=\frac{12}{5}\]which would give you the answer for T if the POSITION attained a relative extreme at T seconds.

Answer 16

Normally it's a very, very unwise decision to assume a question is wrong, but in this case it looks like you've hit the jackpot. Sorry I couldn't be of more help!