$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

$y - y_1 = m(x - x_1)$

$m = f'(x_0)$, where $x_0$ is the x-coordinate of the point of tangency.

1. Find $f'(x)$. 2. Evaluate $f'(x_0)$ to find the slope $m$. 3. Use point-slope form: $y - f(x_0) = m(x - x_0)$.

1. Plug into the limit definition: $\lim_{h \to 0} \frac{(x+h)^2 - x^2}{h}$. 2. Expand: $\lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h}$. 3. Simplify: $\lim_{h \to 0} \frac{2xh + h^2}{h}$. 4. Factor and cancel: $\lim_{h \to 0} (2x + h)$. 5. Evaluate the limit: $2x$.

1. Find $f'(x)$: $f'(x) = 2x$. 2. Find the slope at $x = 1$: $f'(1) = 2$. 3. Find the y-coordinate at $x = 1$: $f(1) = 1$. 4. Use point-slope form: $y - 1 = 2(x - 1)$.

1. Find $f'(x)$: $f'(x) = 3x^2 - 3$. 2. Set $f'(x) = 0$: $3x^2 - 3 = 0$. 3. Solve for $x$: $x = \pm 1$.

1. Find a common denominator: $\lim_{h \to 0} \frac{\frac{x - (x+h)}{x(x+h)}}{h}$. 2. Simplify: $\lim_{h \to 0} \frac{\frac{-h}{x(x+h)}}{h}$. 3. Multiply by the reciprocal: $\lim_{h \to 0} \frac{-h}{x(x+h)h}$. 4. Cancel $h$: $\lim_{h \to 0} \frac{-1}{x(x+h)}$. 5. Evaluate the limit: $\frac{-1}{x^2}$.

1. Plug into the limit definition: $\lim_{h \to 0} \frac{[3(x+h)^2 + 4(x+h)] - [3x^2 + 4x]}{h}$. 2. Expand: $\lim_{h \to 0} \frac{3x^2 + 6xh + 3h^2 + 4x + 4h - 3x^2 - 4x}{h}$. 3. Simplify: $\lim_{h \to 0} \frac{6xh + 3h^2 + 4h}{h}$. 4. Factor and cancel: $\lim_{h \to 0} (6x + 3h + 4)$. 5. Evaluate the limit: $6x + 4$.

$f(x)$ is increasing.

$f(x)$ is decreasing.

$f(x)$ has a horizontal tangent, which could be a local max, local min, or a stationary point.

$f(x)$ is concave up.

$f(x)$ is concave down.

$f'(x) = 0$ (a horizontal line at y=0).

$f'(x)$ does not exist at that point (non-differentiable).

The graph of $f'(x)$ is a horizontal line, representing the constant slope of $f(x)$.