calculus question..should be easy for most of you mathematicians.?

Question

Hey guys thanks for this!

I'm just having a hard time with this question.

Find the coordinates and nature of the turning point(s) of the curve:
y = x^3/3 - x^2 - 8x + 3.

thank you soo much for this!!

10 pts for most detailed..not that detailed..just detailed enough that you can actually manage to finish the equation.

Answer 1

find the derivative and set it equal to 0

y' = x^2 - 2x - 8


0 = x^2 - 2x - 8

0 = (x - 4) (x + 2)

x = 4 or -2

plug 0 back in for x

and the answer should be (4,-23.666.. and (-2,12.333..)

Answer 2

It has been several years since I have taken calculus, but
here is what I remember. This problem involves taking 1st and 2nd derivatives. The first dirivative is essentially the slope of the line. When the slope = 0 you either have a min or max or point of inflection. The second derivative tells you whether the point in question is indeed a max, min or a point of inflection. So here goes my solution,
y=x^3/3-x^2-8x+3.
To find the extrema which is just means the max, min or points of inflection we take the first derivative
denoted by y' and set = 0 and solve for x.
Taking the first derivative,
y' =x^2-2x-8=0
(this is the product rule you can look it up in your text if you have any questions). To solve for x we factor the quadratic equation
(x-4)(x+2) = 0
This has 2 solutions x= 4, and x= -2
Now we apply the 2nd derivative test. We take the 2nd derivative and evaluate it at the extreama x=4, and x=-2.
If the second derivative is positive, the then extreama is a min. if the second derivative is negative, then the extreama is
a max, and if the extrema = 0 then it is a point of inflection.
Taking the 2nd derivative
y''(x) = 2x-2
Now we evaluate it at our two extreama x=4 and x= -2
y''(4) = 2*4-2 = 8-2 = 6 >0, so x=4 is a minimum.
y''(-2) =2*(-2)-2 =-4-2 = -6 , so x=-2 is a maximum.

.

Woops I just noticed that I forgot the question asks for coordinates ie x,y values. We have the x values x=4
plugging back into the original eqn.
y(x) = x^3/3-x^2-8x+3
y(4) = 4^3/3-4^2-8(4) +3 = 64/3-16-32+3 = 64/3 -45 =64/3-135/3 =-71/3

y(-2) = (-2)^3/3-(-2)^2-8(-2)+3 = -8/3-4+16+3 = -8/3 +15
=45/3-8/3 =37/3
so
So we have your answer the coordinates of turning points are (x,y) = (4,-71/3) is a minimum and (x,y) =(-2,37/3) is a maximum
Hope this helps.
Cool I just notice my answer matches with the top contributor
not bad for a guy who has not picked up a calculus book in over 10 years.
Jessie
the nature

Answer 3

y' = x^2 - 2x - 8
set y' = 0, to find the turning pts.
0 = x^2 - 4x + 2x - 8
0 = (x - 4)(x + 2)
x = 4, x = -2

now find the 2nd derivative to check the nature of the points,
y'' = 2x - 2
y''(4) = 6 ----> y'' is +ve ----> so x = 4 is a minima.
y''(-2) = -6 -----> y'' is -ve -----> so x = -2 is a maxima.

y(4) = 64/3 - 16 - 32 + 3 = 64/3 - 45 = -71/3
(4, -71/3) is the coordinate of the relative minima and the minimum value of y.

y(-2) = -8/3 - 4 + 16 + 3 = 37/3
(-2, 37/3) is the coordinate of the relative maxima and the maximum value of y.

Answer 4

Turning points are where the first derivative equals zero, so

d/dx = x^2 - 2x - 8 = 0
factoring gives (x-4)(x+2)

so the turning points occur where x=-2 or +4

Now put these values back into the original equation to find the y-values of the turning points:

y = (-2)^3/3 - (-2)^2-8(-2)+3 = 37/3
y = (4)^3/3 - (4)^2-8(4)+3 = -71/3

So the points are (-2, 37/3) and (4, -71/3)

Answer 5

first derivative gives you relative max and min, when y' = 0:
y' = x² - 2x - 8
0 = (x-4)(x+2)
x = 4, y = 4^3/3 - 4² - 8(4) + 3 = -23 2/3
x = -2, y = (-2)^3/3 - (-2)² - 8(-2) + 3 = 12 1/3

second derivative tells you what's going on at those points:
y" = 2x - 2
at x = 4, -2(4) - 2 = 6. positive, local min, graph turning up.
at x = -2, 2(-2) - 2 = -6, negative, local max, graph turning down.

when y" = 0, 2x - 2 = 0, 2x = 2, x = 1, that's the inflection point where the graph changes from concave down left of the point to concave up on the right.

Answer 6

y = x^3/3 - x^2 - 8x + 3

dy/dx = x^2 - 2x - 8
d^2y/dx^2 = 2x - 2

At the turning points, dy/dx = 0 and d2/dx^2≠0

x^2 - 2x - 8 = 0
x^2 -4x + 2x - 8 = 0
x(x-4)+2(x-4)=0
(x+2)(x-4) = 0
x = {-2, 4}

2*(-2) - 2 ≠ 0
2*4 - 2 ≠ 0
Both values of x correspond to turning points.

The x-coordinates of the turning points have been determined

Use the original equation:
y = x^3/3 - x^2 - 8x + 3
to determine the y-coordinates

y = {-8/3 - 4 + 16 + 3, 64/3 - 16 - 32 + 3}
= {37/3, -71/3}

Coordinates of turning points:
(-2, 37/3) and (4, -71/3)

Answer 7

The turning points of
y = x^3/3 - x^2 - 8x + 3
are the points where the slope of the function changes inflection.
y' = x^2 - 2x - 8
y'' = 2x - 2 = 0
at the turning point (only one for this function)
x = 1
y = 1/3 - 1 - 8 + 3
(x,y) = (1, - 5 2/3)

Answer 8

y = x^3/3 - x^2 - 8x + 3.
*first multiple all by 3
3y=x^3 - 3x^2 - 24x + 9