   Chapter 3.1, Problem 69E

Chapter
Section
Textbook Problem

Find a cubic function y = ax3 + bx2 + cx + d whose graph has horizontal tangents at the points (–2, 6) and (2, 0).

To determine

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

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)=nxn1

(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+4b2c+d

Therefore, the equation is 8a+4b2c+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),

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

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

y(x)=a(3x31)+b(2x21)+c(1x11)=a(3x2)+b(2x)+c(1)=3ax2+2bx+c

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

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