   Chapter 3.3, Problem 60E

Chapter
Section
Textbook Problem

# Show that the curve y = 1 + x 1 + x 2 has three points of inflection and they all lie on one straight fine.

To determine

To show:

The curve y=1+x1+x2 has three points of inflection, and they lie on one straight line

Explanation

1) Concept:

i) If f''x>0 then fx is concave upward.

ii) If f''x<0 then fx is concave downward.

iii) Inflection point is a point where the concavity of function changes from upward to downward or downward to upward, that is f"(x)=0

2) Given:

y=1+x1+x2

3) Calculation:

Let,

y=f(x)=1+x1+x2

Differentiate f(x) with respect to x,

f'x=ddx1+x1+x2

=1+x2ddx1+x-1+xddx1+x21+x22

=1+x2-1+x2x1+x22

=1+x2-2x-2x21+x22

=1-2x-x21+x22

Again differentiate f'(x),

f"(x)=ddx1-2x-x21+x22

=1+x22ddx1-2x-x2-1-2x-x2ddx1+x221+x24

=1+x22-2-2x-1-2x-x221+x2(2x)1+x24

=1+x2-2-2x-4x1-2x-x21+x23

As the inflection point lies on the curve therefore, f"(x)=0

1+x2-2-2x-4x1-2x-x21+x23=0

1+x2-2-2x-4x1-2x-x2=0

By simplification,

-2-2x-2x2-2x3-4x+8x2+4x3=0

2x3+6x2-6x-2=0

2x3+3x2-3x-1=0

x3+3x2-3x-1=0

Solve for x,

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