Relative Extrema

Question

 list the relative extrema and inflection points for f(x)=(x³/3)+6x²+1

Answer 1

To find the relative extremas, set the derivative of f(x) equal to zero.

 

f'(x) = 0

 

x² + 12x = 0

 

x(x+ 12) = 0

 

 

Setting factors equal to zero

 

x = 0            x + 12 = 0

 

                    x = -12

                 

The x values here are your critical points.  They are the location of the relative extremas.  Next, we use tests points and use them to evaluate the derivative.  You only have relative extremas if the derivative changes signs.

 

Evaluate  f'(-13) , f'(-6) ,  and f'(1).

 

f'(-13) = -13(-13 + 12)

          = -13(-1)

          = 13

 

f'(-6) = -6(-6 + 12)

        = -6(6)

        = -36

 

f'(1) = 1(1 + 12)

       = 13

 

 

Since the derivative changes signs at all of these points, we have a maximum at x=-12 and a minimum at x=0.   Now just evaluate f(-12) and f(0) to get your values.

 

 

 

To find the inflection points, we set the second derivative of f(x) equal to zero.

 

f''(x) = 0

 

2x + 12 = 0

 

2x = -12

 

  x = -6

 

 

This is the only inflection point.  An inflection point is the location where the graph of f(x) changes concavity.   Now evaluate f(-6) to find the y value of the inflection point.

Answer 2

f(x)=(x3/3)+6x2+1

 

f'(x) = x² + 12 x = x(x +12)

 

Setting f'(x) = 0 give the following critical values

x =0, x = -12

 

At this point, you can use the slope test to determine the relative extremum at each critical value.

 

Alternatively, you can use the Second Derivative Test.

There is a relative minimum at x = c if f"(c) > 0 and a relative maximum at x = c if f"(c) < 0

f"(x) = 2x + 12

f"(0) = 0 + 12 > 0 ⇒ At x =0, there is a relative minimum.

f"(12) = 2(-12) + 12 < 0 ⇒ At x = -12, there is a relative maximum.

 

The point of inflection is obtained by setting f"(x) = 0 and solve for x.

Answer 3

take the first derivative: x² + 12x, set this to zero... solve for all possible values of x will provide the local min and max, why? you ask, because that is where the slope is zero for the curve.

 

x = 0 ; x = -12 

 

solve for y for each point: (0, 1) and (-12, a real big number!) both are local extrema....the first is local minimum, second is local maximum.

 

second derivative. 2x + 12; set to zero, solve for x provides us the inflection point, why?, you ask, becasue that is where the slope is changing from direction.. from a positive to a negative or vice verse. So second derivative gets us 2x + 12 = 0; x = -6, the point along x which gives us the inflection point. but I also knew that because the inflection point is half way between 0 and -12.  so the inflection point is (-6, again big number). to lazy to solve it, sorry.

 

I hope this is right! but I think so.