Show \( x^2 + 2x + 2 > 0 \) for all \( x \).

Method 1: Completing the Square

\[ x^2 + 2x + 2 = (x^2 + 2x + 1) + 1 = (x + 1)^2 + 1 \]

Since \( (x + 1)^2 \geq 0 \), we have \( (x + 1)^2 + 1 \geq 1 > 0 \). Thus, \( x^2 + 2x + 2 > 0 \) for all \( x \in \mathbb{R} \).

Method 2: Discriminant

For the quadratic \( x^2 + 2x + 2 \), with \( a = 1 \), \( b = 2 \), \( c = 2 \):

\[ \Delta = b^2 – 4ac = 2^2 – 4 \cdot 1 \cdot 2 = 4 – 8 = -4 \]

Since \( \Delta < 0 \) and \( a = 1 > 0 \), the quadratic opens upward and has no real roots, so \( x^2 + 2x + 2 > 0 \) for all \( x \).

Answer: \( x^2 + 2x + 2 > 0 \) for all \( x \).

Find where \( f \) is increasing, i.e., where \( f'(x) > 0 \):

\[ f'(x) = \frac{2x + 2}{x^2 + 2x + 2} \]

From (a)(i), the denominator \( x^2 + 2x + 2 > 0 \) for all \( x \). Thus, the sign of \( f'(x) \) depends on the numerator:

\[ 2x + 2 > 0 \implies 2x > -2 \implies x > -1 \]

Verify: At \( x = 0 > -1 \), \( f'(0) = \frac{2 \cdot 0 + 2}{0 + 0 + 2} = \frac{2}{2} = 1 > 0 \). At \( x = -2 < -1 \), \( f'(-2) = \frac{2(-2) + 2}{4 – 4 + 2} = \frac{-2}{2} = -1 < 0 \).

Answer: \( f \) is increasing for \( x > -1 \).

Find \( x \) where \( f'(x) = 0 \):

\[ \frac{2x + 2}{x^2 + 2x + 2} = 0 \]

Since the denominator is never zero, set the numerator to zero:

\[ 2x + 2 = 0 \implies x = -1 \]

Answer: \( x = -1 \)

Show \( f”(x) = \frac{-2x^2 – 4x}{(x^2 + 2x + 2)^2} \). Use the quotient rule on:

\[ f'(x) = \frac{2x + 2}{x^2 + 2x + 2} \]

Let \( u = 2x + 2 \), \( u’ = 2 \), \( v = x^2 + 2x + 2 \), \( v’ = 2x + 2 \). Quotient rule:

\[ f”(x) = \frac{u’ v – u v’}{v^2} \]

Numerator: \( u’ v – u v’ = 2 (x^2 + 2x + 2) – (2x + 2)(2x + 2) \)

\[ = 2x^2 + 4x + 4 – (4x^2 + 8x + 4) = 2x^2 + 4x + 4 – 4x^2 – 8x – 4 = -2x^2 – 4x \]

Denominator: \( v^2 = (x^2 + 2x + 2)^2 \)

\[ f”(x) = \frac{-2x^2 – 4x}{(x^2 + 2x + 2)^2} \]

Answer: \( f”(x) = \frac{-2x^2 – 4x}{(x^2 + 2x + 2)^2} \)

Justify that \( x = -1 \) is a local minimum. From (b)(i), \( f'(-1) = 0 \). Evaluate the second derivative:

\[ f”(-1) = \frac{-2(-1)^2 – 4(-1)}{((-1)^2 + 2(-1) + 2)^2} = \frac{-2(1) + 4}{(1 – 2 + 2)^2} = \frac{-2 + 4}{1^2} = \frac{2}{1} = 2 \]

Since \( f”(-1) = 2 > 0 \), the second derivative test confirms a local minimum at \( x = -1 \).

Answer: \( x = -1 \) corresponds to a local minimum since \( f”(-1) = 2 > 0 \).

Given \( f(2) = 3 + \ln 10 \), find \( f(x) \). Integrate:

\[ f'(x) = \frac{2x + 2}{x^2 + 2x + 2} = \frac{2(x + 1)}{(x + 1)^2 + 1} \]

Substitute \( u = x + 1 \), \( du = dx \):

\[ f'(x) = \frac{2u}{u^2 + 1} \]

\[ \int \frac{2u}{u^2 + 1} \, du = \ln(u^2 + 1) + C \]

Back-substitute: \( f(x) = \ln((x + 1)^2 + 1) + C = \ln(x^2 + 2x + 2) + C \).

Use \( f(2) = 3 + \ln 10 \):

\[ f(2) = \ln(2^2 + 2 \cdot 2 + 2) + C = \ln(4 + 4 + 2) + C = \ln 10 + C \]

\[ \ln 10 + C = 3 + \ln 10 \implies C = 3 \]

Answer: \( f(x) = 3 + \ln(x^2 + 2x + 2) \)

Find the normal to \( f \) at \( (2, 3 + \ln 10) \). Slope of the tangent:

\[ f'(2) = \frac{2 \cdot 2 + 2}{2^2 + 2 \cdot 2 + 2} = \frac{6}{10} = \frac{3}{5} \]

Normal slope = negative reciprocal = \( -\frac{5}{3} \).

Point-slope form at \( (2, 3 + \ln 10) \):

\[ y – (3 + \ln 10) = -\frac{5}{3} (x – 2) \]

\[ y – 3 – \ln 10 = -\frac{5}{3} x + \frac{10}{3} \]

\[ y = -\frac{5}{3} x + \frac{10}{3} + 3 + \ln 10 = -\frac{5}{3} x + \frac{19}{3} + \ln 10 \]

Answer: \( y = -\frac{5}{3} x + \frac{19}{3} + \ln 10 \)