Questions: If f(x) = ln (x^3 + 5x^2 + 5) then for what value of x does the graph of f(x) have a relative minimum? A. x = -4 B. x = -2 C. x = 0 D. x = 2

Transcript text: If $f(x)=\ln \left(x^{3}+5 x^{2}+5\right)$ then for what value of $x$ does the graph of $f(x)$ have a relative minimum? A. $x=-4$ B. $x=-2$ C. $x=0$ D. $x=2$

Solution

Solution Steps

To find the value of $ x $ where the graph of $ f(x) = \ln(x^3 + 5x^2 + 5) $ has a relative minimum, we need to follow these steps:

Compute the first derivative $ f'(x) $ and set it to zero to find critical points.
Compute the second derivative $ f''(x) $ and evaluate it at the critical points to determine concavity.
Identify the critical point where $ f''(x) > 0 $, indicating a relative minimum.

Step 1: Compute the First Derivative

We start with the function $ f(x) = \ln(x^3 + 5x^2 + 5) $. The first derivative is given by: \[ f'(x) = \frac{3x^2 + 10x}{x^3 + 5x^2 + 5} \] To find critical points, we set $ f'(x) = 0 $: \[ 3x^2 + 10x = 0 \] Factoring gives: \[ x(3x + 10) = 0 \] Thus, the critical points are: \[ x = 0 \quad \text{and} \quad x = -\frac{10}{3} \]

Step 2: Compute the Second Derivative

Next, we compute the second derivative $ f''(x) $ to determine the concavity at the critical points: \[ f''(x) = \frac{(6x + 10)(x^3 + 5x^2 + 5) - (3x^2 + 10x)(3x^2 + 10x)}{(x^3 + 5x^2 + 5)^2} \]

Step 3: Evaluate the Second Derivative at Critical Points

We evaluate $ f''(x) $ at the critical points:

For $ x = 0 $: \[ f''(0) = \frac{(6(0) + 10)(0^3 + 5(0)^2 + 5) - (3(0)^2 + 10(0))(3(0)^2 + 10(0))}{(0^3 + 5(0)^2 + 5)^2} = \frac{10 \cdot 5}{5^2} = 2 > 0 \] This indicates a relative minimum at $ x = 0 $.
For $ x = -\frac{10}{3} $: The evaluation shows that $ f''\left(-\frac{10}{3}\right) < 0 $, indicating a relative maximum.