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.
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!!
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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?
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
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?Originally Posted by apcalculus