Calculus problem help!?

f(x) = x^3 - 12x + 1

f ' (x) = 3x^2 - 12

f '' (x) = 6x

equate f ' (x) to zero to get critical points

3x^2 - 12 = 0

=> x^2 = 4

x = -2 and 2

in the interval ( - ∞, -2 ) f is increasing

in the interval (-2, 2) f is decreasing

in the interval (2, ∞) f is increasing

f(-2) = -8 + 24 + 1 = 17 is local maximum

f(2) = 8 - 24 + 1 = -15 is local minimum

equate f '' (x) to zero to get inflection points

6x = 0

=> x = 0

x = 0 is only inflection point

in the interval ( - ∞, 0 ), f '' (-1) = -6 < 0 --> concave down

in the interval (0, ∞), f '' (1) = 6 > 0 --> concave up

f (x) = x^3 - 12x + 1

. . . the first derivative set to 0 finds turning or stationary points

f ' (x) = 3x^2 - 12

3x^2 - 12 = 0

3 * (x + 2) * (x - 2) = 0

x = 2 ... x = - 2

. . . the second derivative evaluated at x = 2 and -2 determines if those points are min, max, or neither.

f ' ' (x) = 6x

f ' ' (2) = 6*2 = 12 <== positive value indicates x=2 is a local minimum

f ' ' (-2) = 6*(-2) = -12 <== negative value indicates x=-2 is a local maximum

a.)

x = - 2 is a maximum, and x=2 is a minimum ... so

x = - infinity to -2 is increasing

x = -2 to +2 is decreasing

x = +2 to + infinity is increasing

b.)

f (-2) = (-2)^3 - 12*(-2) + 1 = 17

f (2) = (2)^3 - 12*(2) + 1 = - 15

c.)

. . . the second derivative set to 0 finds inflection points, or where concavity changes

6x = 0

x = 0 <=== inflection point

x = - 2 is a maximum, so must be concave down

concavity changes at the inflection point(s) ... so

x = - infinity to 0 is concave down

x = 0 to + infinity is concave up

I'll answer b first because it helps you find a. To find the minimum and maximum values of f, you should first derive your function, which should be 3x^2-12. Knowing that a derivative equals the slope of the tangent line, and that minimums and maximums occur when the slope is 0, you should set 3x^2 -12 to zero. This should give you zeroes which are 2 and -2.

Now you should plug in points within intervals: infinity to -2, -2 to 2, and 2 to infinity, into your derived equation 3x^2-12. If the result is positive, your function is increasing. If negative, your function should be decreasing.

For c, know that inflection points are the points where the graph changes concavity. This link discusses concavity: http://www.sosmath.com/calculus/diff/der15/der15.h... To find these, derive your equation again to get 6x. Find your zero, which is simply 0. This is your point of concavity. Plug in points once more from infinity to zero and zero to infinity to find your intervals of concavity. A positive result means it is concaved upward like a u, and a negative result means that your function is concaved downward, like an upside-down u.

a) It will be increasing when the first derivative is positive, and decreasing when the first derivative is negative.

The first derivative is 3x^2 - 12, which is positive when 3x^2 > 12, which is when x^2 > 4, which occurs when x<-2 or x>2

It's negative when 3x^2 < 12, which is when x^2 < 4, which occurs when -2 < x < 2

So, the intervals when it is increasing:

(-inf, -2) U (2, +inf)

Decreasing:

(-2, 2)

b)

There is a local maximum at x=-2, and that maximum is (-2)^3 - 12(-2) + 1 = -8 + 24 + 1 = 17

There is a local minimum at x=2, and that minimum is (2)^3 - 12(2) + 1 = 8 - 24 + 1 = -15

c)

For concavity, we go by the second derivative, which is 6x

When x < 0, it is negative, and thus concave down.

When x > 0, it is positive, and concave up.

a) Take the derivative of f(x).

f'(x) = 3x^2 - 12.

Solve for zero. There should be two points.

3x^2 = 12

x^2 = 4

x = ± 2.

Now - at these points, the slope is 0, and it changes from positive to negative or vice versa.

Find f'(x) at x = -3.

3x^2 - 12

3 * 9 - 12

Positive. Therefore, from negative infinity to -2, f(x) is increasing.

From here, you can say that from -2 to 2, f(x) is decreasing, and from 2 to infinity, f(x) is increasing again.

b) Those two points, ±2, are your minimum and maximum.

So calculate them for f(x).

f(2) = 2^3 - 12 * 2 + 1

= 8 - 24 + 1

=-15

f(-2) = -2^3 + 12 * 2 + 1

= -8 + 24 + 1

= 17

So your two points are (-2, 17) and (2, -15). The first one is your local maximum, and the second one is your minimum.

c) Take the second derivative of f(x).

f''(x) = 6x

Solve for zero. Obviously, here x = 0. This is your inflection point. From negative infinity to 0, your concavity is negative. From 0 to positive infinity, your concavity is positive.

Mike

I can only help on A and B. If you have a graphing calculator plug it in and where you have a positive slope is where it is increasing. Negative slope is where its decreasing. Same thing in the calculator for B. If you have an 83 plus go to graph>2nd> calc> than hit max or min.

The mathematical way to find b is find the derivative which is 3x^2-12 than set it equal to zero. Than solve for x. This will give you the x value for the min and max. Which is plus or minus 2. Plug -2 into the original equation which will give you -15. So the coordinate for that max is (-2, 18). Repeat for positive 2. which is (2,15)

HI there Buddy, :)

a) first u need to find when a change occurs in this graph or in other words u have to find the turning points,

so firstly u find dx/dy for f(x)

f ' (x)= 3x^2 -12 then u make it equal to zero>> 3x^2 -12=0 >> x^2=4 >> x is +2 or -2

what u need to do now is to find the 2nd derivative,

f '' (x)= 6x >> now u put the two answers we found to this one>>

when x=2>>answer is + then it has a maximum

when x=-2>> answer is - then it has a minimum

hope u found it usefull

df/dx=3x^2-12

3x^2-12>0 means function is rising

x^2>4

which is true for

x<-2 and x>2

df/dx=0 solves for min or max (points x=-2 and x=2)

maximum is at x=-2, minimum is at x=2

inflextion points are at

d2f/dx2=0

which is at

6x=0 or x=0