I need help on My Calc HW

OK so i have to find the inflection point, Concave up and concave down using the 2nd Derivative, Increasing and decreasing point using the 1st Derivative.

exp#= exponent to the ___ power so X(exp 2) = X squared

the problem is F(x) = (x-2)/(x{exp2} - x +1){exp2}

I had to use the quotient rule and i got the 1st Derivative (WARNING I am not responsible for brain explosions ):

dx/dy = [(1)(x{exp2} - x +1){exp2}-(2)(2x-1)(x-1)]
(x{exp2} - x +1){exp4}

then i had to get the 2nd Derivative of the function which was....

[(2)(2x-1)-2((2x)(x-2)+2x-1)(1))(x{exp2} - x +1){exp4}-((4)(x{exp2} - x +1){exp3}))(2x-1))((x{exp2} - x +1)-(2)(2x-2)(x-2))]
(x{exp2} - x+1){exp8}

Set the 1st and 2nd Derivative to 0 then solve both to get the... well if you know Calc you know what to do... the teacher said something about the Quadratic formula being involved AND we can't use a calculator so steps HAVE to be shown

Stuff like this is why people hate math. You've done your first derivative wrong - make sure you're doing the chain rule correctly.

 Spoiler

Use quotient rule.

[img]http://www.codecogs.com/eq.latex?f'(x) = \frac{(1)(x^2-x+1)^2-\frac{d}{dx}(x^2-x+1)^2(x-2)}{(x^2-x+1)^4}[/img]

Use chain rule.

[img]http://www.codecogs.com/eq.latex?f'(x) = \frac{(x^2-x+1)^2-2(x^2-x+1)(2x-1)(x-2)}{(x^2-x+1)^4}[/img]

Simplify.

[img]http://www.codecogs.com/eq.latex?f'(x) = \frac{(x^2-x+1)-2(2x-1)(x-2)}{(x^2-x+1)^3}[/img]

[img]http://www.codecogs.com/eq.latex?f'(x) = \frac{(x^2-x+1)-2(2x^2-5x+2)}{(x^2-x+1)^3}[/img]

[img]http://www.codecogs.com/eq.latex?f'(x) = \frac{-3(x^2-3x+1)}{(x^2-x+1)^3}[/img]

At the risk of bad algebra, this should be the first derivative.

Also, your Leibnitz notation is wrong. It's , not .

What are you asking for exactly? The answer itself? I'm not convinced you know what "inflection point", "concave up", and "concave down" even mean.

Stuff like this is why people hate math. You've done your first derivative wrong - make sure you're doing the chain rule correctly.

 Spoiler

Use quotient rule.

[img]http://www.codecogs.com/eq.latex?f'(x) = \frac{(1)(x^2-x+1)^2-\frac{d}{dx}(x^2-x+1)^2(x-2)}{(x^2-x+1)^4}[/img]

Use chain rule.

[img]http://www.codecogs.com/eq.latex?f'(x) = \frac{(x^2-x+1)^2-2(x^2-x+1)(2x-1)(x-2)}{(x^2-x+1)^4}[/img]

Simplify.

[img]http://www.codecogs.com/eq.latex?f'(x) = \frac{(x^2-x+1)-2(2x-1)(x-2)}{(x^2-x+1)^3}[/img]

[img]http://www.codecogs.com/eq.latex?f'(x) = \frac{(x^2-x+1)-2(2x^2-5x+2)}{(x^2-x+1)^3}[/img]

[img]http://www.codecogs.com/eq.latex?f'(x) = \frac{-3(x^2-3x+1)}{(x^2-x+1)^3}[/img]

At the risk of bad algebra, this should be the first derivative.

Also, your Leibnitz notation is wrong. It's , not .

What are you asking for exactly? The answer itself? I'm not convinced you know what "inflection point", "concave up", and "concave down" even mean.

First: Thanks

Second: Well the book said:
Find: (a) the intervals on which F is increasin, (b) the intervals on which F is decresing, © the open intervals on which F is concave up, (d) the open intervals on which F is concave down, and (e) the x-coodinates of all inflection points.

Third i need both the simplified form of the first and 2nd derivative set each = to 0 and solve.. i did that to the one u gave me and I used to quad formula and got

3(+ or -)(square root 5)
2

now the answers for this problem is in the back and that turned (the fraction above) is the answer for A but i'm not 100% sure WHY it's the answer or why b,c,and d are the answers they are... i know E is the solved version of the 2nd derivative (solved that way shown above)

Increasing: the intervals of x where f'(x)>0
Decreasing: the intervals of x where f'(x)<0
Inflection points: intervals of x where f"(x)=0
Concave up: etc. f"(x)>0
Concave down: etc. f"(x)<0

Shouldn't all these definitions be in the book somewhere?

First off, I think you've differentiated it wrong. You've not used the chain rule properly when differentiating the bottom bracket. So let's start again.



Differentiating that by quotient and chain rule we get:

[IMG]http://codecogs.com/eq.latex?f'(x)&=\frac{(x^2-x+1)^2(1)-(x-2)(x^2-x+1)(2)(2x-1)}{(x^2-x+1)^4}[/IMG]

Which I simplify down to:

[IMG]http://codecogs.com/eq.latex?f'(x)=-\frac{3(x^2-3x+1)}{(x^2-x+1)^3}[/IMG]

We might as well get the 2nd derivative while we're here. Again, using quotient and chain rule:

[IMG]http://codecogs.com/eq.latex?f''(x)=\frac{(x^2-x+1)^3(-3[2x-3])-(-3[x^2-3x+1])(x^2-x+1)^2(3)(2x-1)}{(x^2-x+1)^6}[/IMG]

Which simplifies to:

[IMG]http://codecogs.com/eq.latex?f''(x)=\frac{6x(2x^2-8x+5)}{(x^2-x+1)^4}[/IMG]

Now you might want to double-check those derivatives, but I hope they're right lol.

Now for part (a), intervals for which it's an increasing function. As stated above, it's increasing when it's gradient is positive (i.e. f'(x) > 0), so:

[IMG]http://codecogs.com/eq.latex?f'(x)>0[/IMG]




Using the quadratic formula, I get (as you've previously mentioned):





Right, now for part (b), intervals for which it's a decreasing function. Given that we've already done part (a), this is pretty trivial. It's either increasing, decreasing or stationary. Given that the two solutions above are the only stationary points (when f'(x) = 0), the opposite intervals must describe when the function is decreasing:



Part ©, intervals for which the function is concave up. As already said, when f''(x) > 0:

[IMG]http://codecogs.com/eq.latex?f''(x)>0[/IMG]



Either:



Or:



Therefore:



Again, as with part (b), we can get part (d), intervals for which the function is concave down, from the above answer:



Finally, part (e), the co-ordinates of all inflection points. Again, this is trivial given our answers to parts © & (d), as we already know the 3 x-values for which f''(x) = 0:



Hope that's all correct, lol. If it's not, I at least hope that it helps.