precalc paranoia

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bcddd214

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f(x)=3x^4+4x^3

the instruction say "sketch the curve".
My first thought is to just do complete the square, set to zero and see what happens from there.
This is the 'ol math fear complex.
The instruction are so vague I don't know where to begin.... :(
just a point in the right direction and I am sure I can launch...
 
bcddd214 said:
f(x)=3x^4+4x^3

the instruction say "sketch the curve".
My first thought is to just do complete the square, set to zero and see what happens from there.
This is the 'ol math fear complex.
The instruction are so vague I don't know where to begin.... :(
just a point in the right direction and I am sure I can launch...
Click to expand...

This is NOT a complete the square problem.

Can you factorize the expression?

What are the roots (zeros) of the expression?

This is a 4th order expression - what should be the "approximate" shape of the curve?
 
this is why I feel so stoopid.
Factoring just gives
=x^3 (3x+4)
That doesn't do much for me
setting it to zero at this point just doesn't make sense to me.
3x=-4
x=-4/3?

The split/goofy x^3 parabola like thing, offset on the x axis by -4/3?

like

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bcddd214 said:
this is why I feel so stoopid.
Factoring just gives
=x^3 (3x+4)
That doesn't do much for me
setting it to zero at this point just doesn't make sense to me.
3x=-4
x=-4/3?

The split/goofy x^3 parabola like thing, offset on the x axis by -4/3?

like

|
|_
|
|
Click to expand...

For this function - by factoring you found a "zero" at x = 0 and a zero at x = -4/3.

The "zero" at x = 0 is a special kind - it is called a "repeated zero" (in this case 3 zeros at one point x =0). At this point, the curve will be parallel to the x-axis. From your choice of topic - precalc paranoia - I gather that you are not to use calculas to help you in sketching. Your curve will have a local minimum at x=-1 and at it will be flat around x=0.

You can see that for very large value of |x| (positive and negative) y is positive. So it will "kind of look like" a parabola. I suppose you'll have to plot this function by evaluating this function at different values of 'x'. When x>-4/3 and x<0, take small increment x - so that you can catch the "ups and downs" of the curve.
 
Don't hate me for being stoopid, but I catch the -4/3 easy, but why the second zero at zero?
 
I hit the same road block with this one.
f(x)=x^4+2x^2-3
I am thinking
(x^4+2x^2 )-3
?
but that doesn't look correct.
 
I think I am getting dumber by the hour.
f(x)=(x^2+1)/(x^2-2)
I look at it at first and think difference of 2 squares in the denominator.
then I notice the exponent.... step 1 just does come easy for me.
 
and with this one, step one came too easy. I don't think I am on the right track at all.

f(x)=xe^x
f'(x)=e^x
 
f(x)=3x^4+4x^3
the instruction say "sketch the curve".

Factoring just gives
=x^3 (3x+4)
That doesn't do much for me
setting it to zero at this point just doesn't make sense to me.
3x=-4
x=-4/3?
Click to expand...

If we want to sketch the curve, it’s nice to know a few key points, such as x and y intercepts. Since all x-intercepts have a y value of zero, that’s why we set the equation equal to zero – to find x-intercepts. Once we have the function in factored form, we set each individual factor equal to zero and solve. In this case we have two factors: (x^3) and (3x+4). [x^3 is really three factors, but they are all identical.] Therefore we have two equations to solve:

0 = (3x+4) and 0 = x^3

So, our solutions are x = -4/3 and x = 0. Now we know two points on our graph: (-4/3,0) and (0,0). Make sense now?

Since this function is a 4th degree (x^4) polynomial, it’s general shape is a “U”. Since the leading coefficient is positive (3), the “U” opens upward. Subhotosh has already explained the shape of the curve as it passes through the origin.

Second problem: f(x)=x^4+2x^2-3
This is similar to the last problem in that it is another 4th degree poly, so it will have the same general U shape. Again, our first step is to factor the polynomial. Try a “u substitution’. Let u = x^2. Now the function reads as f(u) = u^2 + 2u – 3, which you should know how to factor. Once you have factored it, replace the u’s with x^2, set the factor equal to zero, and solve as we did above.

Third problem: f(x)=(x^2+1)/(x^2-2)
You were on the right track to examine the denominator. Setting the denominator equal to zero will allow you to find vertical asymptotes or holes depending on the function. Tkhunny, et al, have already discussed that with you in other posts. After determining the vertical asymptotes, set the numerator equal to zero to look for x-intercepts. (Hint: there are none in this case.) Next, plug in x values close to your asymptotes (on both sides of the asymptotes) to determine function behavior in those locations. From prior posts, I believe you already know that the horizontal asymptote is y = 1 (the ratio of the leading coefficients, since numerator and denominator are to the same power.)

Third problem:
f(x)=xe^x
f'(x)=e^x
Click to expand...
[Incorrect. You need to review both the product rule and “e” rules of differentiation.]
 
bcddd214 said:
Don't hate me for being stoopid, but I catch the -4/3 easy, but why the second zero at zero?
Click to expand...

Someone else sent me this as the solution.
Is everyone off track with this one?

1st deriv: y’ = 12x^3 + 12x^2 = 12x^2(x+1)
Y’ equals 0 when x=0 or x=-1
???????????????
 
bcddd214 said:
bcddd214 said:
Don't hate me for being stoopid, but I catch the -4/3 easy, but why the second zero at zero?
Click to expand...

Someone else sent me this as the solution.
Is everyone off track with this one?

1st deriv: y’ = 12x^3 + 12x^2 = 12x^2(x+1)
Y’ equals 0 when x=0 or x=-1
???????????????
Click to expand...


2nd deriv:
y’’=36x^2 + 24x
= 12x(3x+2)

y’’(x) equals 0 when x=0 or x= -2/3
 
1st deriv: y’ = 12x^3 + 12x^2 = 12x^2(x+1)
Y’ equals 0 when x=0 or x=-1
Click to expand...

True.

2nd deriv:
y’’=36x^2 + 24x
= 12x(3x+2)

y’’(x) equals 0 when x=0 or x= -2/3
Click to expand...

Also true.

Do you know the significance of this? What do first and second derivatives reveal about the function?

(PS We cannot know what methods/knowledge are available to you if you have not demonstrated them in your work.)
 
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