To determine

To find: The cubic function whose graph has horizontal tangents at the points (−2,6) and (2,0).

Answer

The function is y=316x3−94x+16.

Explanation

Given:

The cubic function is y=ax3+bx2+cx+d (1)

Derivative rules:

(1) Constant multiple rule: ddx(cf)=cddx(f)

(2) Power rule: ddx(xn)=nxn−1

(3) Sum rule: ddx(f+g)=ddx(f)+ddx(g)

Calculation:

Obtain the values of a, b, c and d in the cubic function.

Take the point (−2,6).

Here, x=−2 and y=6 .

Substitute −2 for x and 6 for y in equation (1),

6=a(−2)3+b(−2)2+c(−2)+d6=a(−8)+b(4)+c(−2)+d6=−8a+4b−2c+d

Therefore, the equation is −8a+4b−2c+d=6 (2)

Take the point (2,0).

Here, x=2 and y=0.

Substitute 2 for x and 0 for y in equation (1).

0=a(2)3+b(2)2+c(2)+d0=a(8)+b(4)+c(2)+d0=8a+4b+2c+d

Therefore, the equation is 8a+4b+2c+d=0 (3)

Obtain the derivative of y=ax3+bx2+cx+d.

The first derivative of y=ax3+bx2+cx+d is

y′(x)=ddx(ax3+bx2+cx+d)

Apply the sum rule (3) and the constant multiple rule (1),

y′(x)=ddx(ax3)+ddx(bx2)+ddx(cx)+ddx(d)=addx(x3)+bddx(x2)+cddx(x)+ddx(d)

Since the derivative of constant function is zero, ddx(d)=0.

y′(x)=addx(x3)+bddx(x2)+cddx(x)+0

Apply the power rule (2) and simplify the terms,

y′(x)=a(3x3−1)+b(2x2−1)+c(1x1−1)=a(3x2)+b(2x)+c(1)=3ax2+2bx+c

Therefore, the first derivative of y is y′(x)=3ax2+2bx+c.

Note that, if the graph of the function f(x) has horizontal tangent at the point (x0,y0) then f′(x0)=0.

Since the graph of the function has horizontal tangents at (−2,6) and (2,0), y′(−2)=0 and y′(2)=0.

Substitute −2 for x in y′(x)=3ax2+2bx+c,

f′(−2)=3a(−2)2+2b(−2)+c=3a(4)−4b+c=12a−4b+c

Since f′(−2)=0, then 12a−4b+c=0 (4)

Substitute 2 for x in f′(x)=3ax2+2bx+c,

f′(2)=3a(2)2+2b(2)+c=3a(4)+4b+c=12a+4b+c

Since f′(2)=0, then 12a+4b+c=0 (5)

Subtract equation (5) from (4) as follows,

−8b=0b=0

Thus, the value of b=0 .

Add equation (2) and (3),

8b+2d=6

Substitute 0 for b,

8(0)+2d=62d=6d=3

Thus, the value of d=3 .

Substitute 0 for b in equation (5) as follows,

12a+c=0c=−12a

Substitute −12a for c ,0 for b and 3 for d in equation (3),

8a+4(0)+2(−12a)+3=08a−24a+3=0−16a=−3a=316

Thus, the value of a=316

Substitute 316 for a in c=−12a,

c=−12a=−12(316)=−94

Thus, the value of c=−94.

Therefore, the values of a, b, c and d are a=316,b=0,c=−94 and d=3.

Substitute the values of a, b, c and d in equation (1),

y=(316)x3+(0)x2+(−94)x+3=316x3+0−94x+3=316x3−94x+3

Therefore the cubic function is y=316x3−94x+16.