Introduction to Optimization Problems - AP Calculus AB

$f'(x) = -\text{sin}(x)$. Derivative of cosine function is negative sine function.

$\frac{v'u - uv'}{v^2}$ for $u/v$. Quotient rule: $ (\text{bottom})(\text{top}') - (\text{top})(\text{bottom}') $ over $ (\text{bottom})^2 $.

$f'(x) = 2\sin(x)\cos(x)$. Using chain rule: $2\sin(x) \cdot \cos(x)$.

$f'(x) = 0$ and $f''(x) > 0$. First derivative zero ensures extremum, second derivative positive confirms minimum.

$f'(x) = \frac{1}{x}$. Natural logarithm derivative is reciprocal function.

Identify the quantity to be optimized. Defining the objective function is essential before analyzing constraints.

$f'(x) = 3x^2$. Power rule applied: $3x^{3-1} = 3x^2$.

To determine concavity and identify local extrema. Second derivative determines whether critical points are maxima or minima.

If $u(x)$ and $v(x)$, then $uv' + vu'$. Product rule: derivative of first times second plus first times derivative of second.

Whether a critical point is a local max or min. Analyzes sign changes of $f'(x)$ around critical points.

$x \text{ in } [0, 3]$. Domain restriction defines the feasible region for optimization.

Critical points at $x = 0$ and $x = 2$. Setting $f'(x) = 3x^2 - 6x = 3x(x-2) = 0$.

$f(x)$ is increasing. Positive derivative means function has positive slope.

It represents the maximum or minimum value. Parabola vertex occurs at the critical point of quadratic function.

If $y = f(g(x))$, then $y' = f'(g(x))g'(x)$. Chain rule: derivative of outside function times derivative of inside function.

$f'(x) = \frac{1}{x}$. Using chain rule: $\frac{1}{5x} \cdot 5 = \frac{1}{x}$.

$f(x)$ is concave down. Negative second derivative indicates downward concavity.

A critical point and the second derivative. Need $f'(c) = 0$ and $f''(c) \neq 0$ to apply test.

$f'(x) = \text{sec}^2(x)$. Derivative of tangent function is secant squared.

To evaluate endpoints for absolute extrema. Domain endpoints must be checked for absolute extrema.

Critical point at $x = 2$. Setting $f'(x) = 2x - 4 = 0$ gives $x = 2$.

$f'(x) = 2x + \text{cos}(x)$. Sum rule: derivative of sum equals sum of derivatives.

$f'(x) = 2x$. Power rule: derivative of $x^n$ is $nx^{n-1}$.

The function to be maximized or minimized. The function being optimized in an optimization problem.

$f'(x) = 0$ and $f''(x) < 0$. First derivative zero ensures extremum, second derivative negative confirms maximum.

$f'(x) = \cos(x)$. Derivative of sine function is cosine function.

$f'(x) = \text{e}^x + x\text{e}^x$. Using product rule: $(1)(e^x) + (x)(e^x) = e^x(1 + x)$.

To identify potential maxima or minima. Critical points are candidates for local extrema.

$f'(x) = -\frac{1}{x^2}$. Using power rule: $x^{-1}$ becomes $-1 \cdot x^{-2}$.