Cubic Function

Question

Find the cubic function of the form f (x) = ax^3 + bx^2 + cx + d, where a ≠ 0 and the coefficients a,b,c,d are real numbers, which satisfies the conditions given below.

Relative maximum: 3,0 ; Relative minimum: 5,−2 ; Inflection point: 4,−1

Answer 1

Set f'(x)=0

 

3ax² + 2bx + c = 0

 

The plug in the values of the critical points in the derivative equation.

 

3a(3)² + 2(b)(3) + c = 0

27a + 6b + c = 0                      eq1

 

3a(5)² + 2(b)(5) + c = 0

75a + 10b + c = 0                     eq2

 

 

Now we plug in the value of the max and min points into the original equation

 

a(3)³ + b(3)² + c(3) + d = 0

27a + 9b + 3c + d = 0                  eq3           (maximum)

 

 

a(5)³ + b(5)² + c(5) + d = -2

125a + 25b + 5c + d = -2              eq4          (mimimum)

 

 

Next, we take the second derivative to deal with the inflection point.

 

6ax + 2b = 0

 

 

Plug in the inflection points values into the second derivative and original equations.

 

6a(-4) + 2b = 0

-24a + 2b = 0                                 eq5

 

a(-4)³ + b(-4)² + c(-4) + d = -1

-64a + 16b - 4c + d = -1                eq6

 

 

As you can see, the 6 bolded equations are your system of equations.  Use them to solve for a, b, c, and d.  I would solve them using a program, rather by hand.

Answer 2

Hi Kayla.

 

I got stuck when all I used was the information from the derivatives.  

 

So I started over and plugged in the x- and y- values given for the min, max and poi into the original function and got a system of 3 equations.

 

From the poi in f'', I found that b = -12a

Then from both of the extrema equations I found that c = 45a

 

When I plugged the maximum (3,1) back into the original f(x), I got 

27a + 9b + 3c + d = 0

 

Into this I put b and c in terms of a.

 

27a + 9(-12a) + 3(45a) + d = 0

 

Simplifying and combining like terms, I got 54a + d = 0, so I found d in terms of a also, 

d= -54a

 

I could then plug b, c, and d (all in terms of a) back into the original function using the x- and y- coordinates of the minimum.

 

So using x=5 and y = -2, the original equation becomes

 

f(5) = 125a + 25b + 5c + d = -2

 

Replacing b, c, and d with their "a-equivalents", I get

 

125a + 25(-12a) + 5(45a) + (-54a) = -2

 

The LHS simplifies down to -4a = -2, so a = (1/2).

 

Plugging a back in to the equations for b, c, and d,

 

b = -12(1/2) = -6

 

c = 45(1/2) = 45/2

 

d = -54(1/2) = -27

 

 

So your final cubic is:

 

f(x) = (1/2)x³ - 6x² + (45/2)x - 27

 

 

Checking this, find that

 

f'(x) = (3/2)x² - 12 x + (45/2)

 

Setting it = 0 and solving the quadratic - first mult everything by 2 to get rid of the fractions

 

 3x² - 24x + 45 = 0

 

Divide everything by 3

 

 x² - 8x + 15 = 0  which easily factors into

 

(x-3)(x-5) = 0,  so min and max are located at x=3 and x=5

 

Now check to see that we get the correct y-values

 

f(3) = 0  (so that works)

f(5) = -2 (so that works)

 

and last, check the point of inflection

 

f''(x) = 3x - 12 and set it = 0

 

3x = 12, so x = 4.

 

Check to see that get correct y-value

 

f(4) = -1 (so that works)

 

Conclusion:  the correct cubic function is

 

 

f(x) = (1/2)x³ -6x² + (45/2)x - 27

Answer 3

HINTS: You have to solve for a, b and c such that

 

f'(x)= 3ax² + 2bx + c = 0 at x = 3 or 5  (critical points for relative extrema)

and f(3) = 0 and f(5) = -2 and

  

f"(x) = 6ax +2b = 0 at x = 4 (inflection point).

 

Can you proceed from here? don't forget to check your work, at the end, when you have resolved a, b & c, to see that the stated conditions do indeed prevail.