# Thread: Finding critical points and equations of tangents.

1. ## Finding critical points and equations of tangents.

1. Determine the critical points (Clasify as maxima, minima), Inflection points and trace the y=16x+4x^2-x^4
2. Find the equation of tangent to the circle x^2 + y^2 = 9 and parallel to 2x + y= 10.

2. Originally Posted by banana_banana
1. Determine the critical points (Clasify as maxima, minima), Inflection points and trace the y=16x+4x^2-x^4
2. Find the equation of tangent to the circle x^2 + y^2 = 9 and parallel to 2x + y= 10.

2.
First find a slope form for the given circle by implicitly differentiating with respect to x:

$\displaystyle 2x + 2y y' = 0$

and solve for y':

$\displaystyle y' = \frac{-x}{y}$

Because the slopes are equal, the derivative form above, when evaluated at the point of tangency, should be equal to the slope of the given line, which is is -2. Note:
$\displaystyle y = -2x+10$

$\displaystyle \frac{-x}{y} = -2$

which means
$\displaystyle x=2y$

Substitute this condition into the equation of the circle to solve for x, then find y by taking half of x.

1.y=16x+4x^2-x^4

Differentiate
$\displaystyle y' = 4x^3- 8x + 16 = 4 (x^3 - 2x + 4)= 4 (x+2) (x^2-2x+2)$

Now solve for zeros (or critical numbers to f(x)) then study the sign to determine types of extrema.

Good luck!!

3. ## Finding the critical points

Originally Posted by apcalculus
2.

1.y=16x+4x^2-x^4

Differentiate
$\displaystyle y' = 4x^3- 8x + 16 = 4 (x^3 - 2x + 4)= 4 (x+2) (x^2-2x+2)$

Now solve for zeros (or critical numbers to f(x)) then study the sign to determine types of extrema.

Good luck!!
, Why is that using using quadratic formula and completing square in getting the CPs would not yield the same value such as this:
Completing Square:
0=x²-2x+2

0= x²-2x+1=-2+1
(x-1)²=-1
x-1=±√-1
x=1±√-1

-1±√(-2)²-4(1)(2)
(2)(1)
-1±√-4
2

4. Originally Posted by banana_banana
, Why is that using using quadratic formula and completing square in getting the CPs would not yield the same value such as this:
Completing Square:
0=x²-2x+2

0= x²-2x+1=-2+1
(x-1)²=-1
x-1=±√-1
x=1±√-1 Mr F says: Do you realise that this value of x is not real?

-1±√(-2)²-4(1)(2) Mr F says: b = -2, not -1 (which is what you substituted in the bit of have highlighted).
(2)(1)
-1±√-4
2
Your equations would look much better and be easier to post if you learnt some basic latex: http://www.mathhelpforum.com/math-help/latex-help/

5. Originally Posted by banana_banana
, Why is that using using quadratic formula and completing square in getting the CPs would not yield the same value such as this:
Completing Square:
0=x²-2x+2

0= x²-2x+1=-2+1
(x-1)²=-1
x-1=±√-1
x=1±√-1

-1±√(-2)²-4(1)(2)
(2)(1)
-1±√-4
2
When you apply the quadratic formula, b = -2, so -b is 2. For some reason you start with a -1 in the numerator in your work.

Pull out the 4 from the square root as a 2, and you get:

$\displaystyle \frac{2 \pm 2 \sqrt{-1}}{2} = 1 \pm i$

Did you need the complex zeros?

6. ## Critical Points and Equation of tangent Circle

so from that I can conclude that from X+2=0, i can have X= -2, and for 1±√-1, how should I have this If it has plus and minus sign, how could i fit it in the parenthesis because both will yield a different value of y.

First Critical Point (-2, -48)
SEcond critical point (1±√-1,?)

TANGENT EQUATION OF CIRCLE
Originally Posted by apcalculus
First find a slope form for the given circle by implicitly differentiating with respect to x:

$\displaystyle 2x + 2y y' = 0$

and solve for y':

$\displaystyle y' = \frac{-x}{y}$

Because the slopes are equal, the derivative form above, when evaluated at the point of tangency, should be equal to the slope of the given line, which is is -2. Note:
$\displaystyle y = -2x+10$

$\displaystyle \frac{-x}{y} = -2$

which means
$\displaystyle x=2y$

Substitute this condition into the equation of the circle to solve for x, then find y by taking half of x.
I dont understand what you mean about this? If i substitute X=2y to the X^2+Y^2=9 , i will get a y=(√9/5). then what will be the next?

7. Originally Posted by banana_banana
so from that I can conclude that from X+2=0, i can have X= -2, and for 1±√-1, how should I have this If it has plus and minus sign, how could i fit it in the parenthesis because both will yield a different value of y.

First Critical Point (-2, -48)
SEcond critical point (1±√-1,?)
Have you realised yet that $\displaystyle 1 \pm \sqrt{-1} = 1 \pm i$ are not real (I did point this out in my earlier post)? Surely the implied domain of $\displaystyle y=16x+4x^2-x^4$ is real numbers.

8. Originally Posted by banana_banana
[snip]TANGENT EQUATION OF CIRCLE
I dont understand what you mean about this? If i substitute X=2y to the X^2+Y^2=9 , i will get a y=(√9/5). then what will be the next?
What you do next is recall that x=2y and substitute the value of y ....

(By the way, please don't edit posts that have been replied to by adding slabs of new stuff. Make a new post in the thread).