AP Calculus AB Exam Question

Let $f(x)$ be a differentiable function defined for all real numbers $x$. The graph of $f(x)$ is shown below:

1. Find the derivative of $f(x)$ at $x = -3$.

Answer with Step-by-Step Detailed Explanation

To find the derivative of $f(x)$ at $x = -3$, we need to determine the slope of the tangent line to $f(x)$ at that point. The derivative of a function at a specific point represents the rate of change of the function at that particular point.

Given that the graph of $f(x)$ is a cubic function with four x-intercepts, we know that its derivative will be a quadratic function. Let's denote the derivative of $f(x)$ as $f'(x)$.

To find $f'(x)$ at $x = -3$, we first need to locate the point on the graph corresponding to $x = -3$. From the graph provided, we can determine that $f(-3) = 0$.

Next, we need to find the slope of the tangent line at this point. We can compute this slope by taking the derivative of $f(x)$ with respect to $x$ and then evaluating it at $x = -3$.

Using the power rule for derivatives, the derivative of $f(x)$ will be the sum of the derivatives of each term. As $f(x)$ is a cubic function, it can be expressed as $f(x) = ax^3 + bx^2 + cx + d$, where $a$, $b$, $c$, and $d$ are constants.

Differentiating each term, we have:

$f'(x) = 3ax^2 + 2bx + c$

Substituting $x = -3$ into this expression, we obtain:

$f'(-3) = 3a(-3)^2 + 2b(-3) + c$

Simplifying,

$f'(-3) = 9a - 6b + c$

Thus, the derivative of $f(x)$ at $x = -3$ is $9a - 6b + c$.

Answer: $f'(-3) = 9a - 6b + c$