Find set of values such that stationary/inflection points exist.

Question

Let f:→x∈ R I x²+bx+1>0}→R defined by f(x)=log_e(x²+bx+1) where b∈R is a constant.

a)Find the set S of values of b for which f has a stationary point. Show your reasoning.

b)Find the set T of values of b for whichfhas two distinct inflection points.

Answer 1

Let's define

g(x) = x² + bx + 1

Then

g'(x) = 2x + b

g"(x) = 2

and

f(x) = ln(g(x))

Let's calculate the first and second derivative of f

f(x) = ln(g(x))

f'(x) = g'(x)/g(x)

f"(x) = g"(x)/g(x) - (g'(x))²/(g(x))²

a)

f has a stationary point for those x where

f(x) is defined, i.e. g(x) > 0 (1) and

f'(x) = 0, i.e., g'(x)/g(x) = 0 (2)

(2) simplifies into

g'(x) = 0

2x + b = 0

x = -b/2

Therefore f has a stationary point for those b where x = -b/2 satisfies (1)

g(-b/2) > 0

(-b/2)² + b(-b/2) + 1 > 0

b²/4 - b²/2 + 1 > 0

b² < 4

-2 < b < 2

S = {b | -2 < b < 2}

b)

Inflexion points are those values of x where

f(x) is defined, see (1), and

f"(x) = 0 (3)

First expand, simplify, and solve (3):

g"(x)/g(x) - (g'(x))²/(g(x))² = 0

g"(x)g(x) - (g'(x))² = 0

2(x² + bx + 1) - (2x + b)² = 0

2x² + 2bx + 2 - 4x² - 4xb - b² = 0

x² + bx + b²/2 - 1 = 0 (4)

x = (-b ± √(b² - 4(b²/2 - 1)))/2 = (-b ± √(4-b²))/2

That yields two distinct solutions provided 4-b² > 0,

which simplifies to

-2 < b < 2 (5)

Secondly any potential inflection point (i.e., any x satisfying (4)) must satisfy (1), which expands into

x² + bx + 1 > 0

That can be rewritten as

x² + bx + b²/2 - 1 - (b²/2 - 1) + 1 > 0

Using (4) this simplifies into

-(b²/2 - 1) + 1 > 0

That is simplifies into (5)

T = {b | -2 < b < 2}