# Thread: [Urgent] help!! calculus, function analysis.

1. ## [Urgent] help!! calculus, function analysis.

Hi, I'm stuck on my calculus problem which is due on tomorrow.

Here's the full question,

Q. For the following functions, do a full analysis of the function itself, the first derivative, and the second derivative.

- domain, intercepts, and vertical and horizontal asymptotes.
Find and classify all critical points
Find all intervals where the function is increasing or decreasing.
Find all intervals of concavity, and inflection points.
Finally, sketch the graph of the function, using the information you obtained.

- Is this mean by I have to do the analysis of three function? since it says, i have to do the analysis of the function itself, the first derivative and second.

i) f(x) = x(x^2+4)^2

I got the first and second derivatives,
f'(x) = (x^2+4)(5x^2+4)
f''(x) = 4x(5x^2+12)

Since I tried by looking at the text book and do the analysis.. I dunno where to start and what to do.. Also, for getting critical points, let say where f(x) = 0, I can't find the roots since i get no real number..

ii) f(x) = x^(5/3) - 5x^(2/3)
i also have the derivatives for these too.
f'(x) = 5/3*x(2/3) - 10/3*x^(-1/3)
f''(x) = 10/9*x^(-1/3) + 10/9*x(-4/3)

for this one ii), I have to care for cusps or vertical tangents..

iii) f(x) = 2cosx+sin2x
also got the derivatives.
f'(x) = -2sin(x)+2cos(2x)
f''(x) = -2cos(x) - 4sin(2x) ==> is this right?

and for this one, I have to care about whether the function is periodic or not..

Please math people.. help me.. I'm a newb

2. (i)The second derivative is $f''(x) =4x(5x^2+12)$
(ii)Everything's nice.
(iii) $\frac{d(sin(u)}{dx} = cos(u)\frac{du}{dx}$
In this case, $u = 2x \implies \frac{du}{dx} = 2$
$f(x) = 2cos(x)+sin(2x) \implies f'(x) = -2sin(x)+2cos(2x)$

3. ## -

Originally Posted by vincisonfire
The second derivative is $*f''(x) =4x(5x^2+12)$
ouch, i forgot to put 4

thanks.

4. Are you sure you did the derivatives, right, I think you might be missing chain rule and product rule on some of them. For example, the derivative of sin(2x) is 2(cos(2x). The derivative of $x(x^2+4)^2$ would be something like $(x^2+4)^2 + ((2x)(x^2+4))^2$, unless I messed up somewhere.

5. ## -

Originally Posted by terr13
Are you sure you did the derivatives, right, I think you might be missing chain rule and product rule on some of them. For example, the derivative of sin(2x) is 2(cos(2x). The derivative of $x(x^2+4)^2$ would be something like $(x^2+4)^2 + ((2x)(x^2+4))^2$, unless I messed up somewhere.
the derivative of sin(2x) is 2(cos(2x)) i guess?

but i think my derivative of x(x^2+4)^2 is right..?

it's (x^2+4)^2 + x*2(x^2+4)(2x)..

since i have to use power rule for the second part to inside too..

6. As I expected, I messed up on that, but there is one solution for when f(x)=0, namely when x = 0.

7. ## -

Originally Posted by terr13
As I expected, I messed up on that, but there is one solution for when f(x)=0, namely when x = 0.
could you help me doing the trig derivatives too?

the first and the second?

and yeah i knew that f(0) = 0, but what do i do with the imaginary roots?.. i'm sorry to say this but i didn't go to lectures.. that's why i'm screwing up what to do.. and also i dunno what should i do with the f(x) = 0 and f'(x) = 0 and stuff..

8. Originally Posted by vincisonfire
(i)The second derivative is $f''(x) =4x(5x^2+12)$
(ii)Everything's nice.
(iii) $\frac{d(sin(u)}{dx} = cos(u)\frac{du}{dx}$
In this case, $u = 2x \implies \frac{du}{dx} = 2$
$f(x) = 2cos(x)+sin(2x) \implies f'(x) = -2sin(x)+2cos(2x)$