Use definition (1) (p. ​133​) to find the slope of the line tangent to the graph of f at P.
f(x) = x² - 5; P(3,4)

Verified step by step guidance

1. Step 1: Recall the definition of the derivative as the slope of the tangent line at a point. The derivative of a function \( f(x) \) at a point \( x = a \) is given by the limit \( f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \).
2. Step 2: Identify the function \( f(x) = x^2 - 5 \) and the point \( P(3, 4) \). We need to find the derivative \( f'(x) \) and evaluate it at \( x = 3 \).
3. Step 3: Substitute \( f(x) = x^2 - 5 \) into the derivative definition: \( f'(x) = \lim_{h \to 0} \frac{(x+h)^2 - 5 - (x^2 - 5)}{h} \).
4. Step 4: Simplify the expression inside the limit: \( (x+h)^2 - 5 - (x^2 - 5) = x^2 + 2xh + h^2 - 5 - x^2 + 5 = 2xh + h^2 \).
5. Step 5: Factor out \( h \) from the numerator: \( \frac{2xh + h^2}{h} = \frac{h(2x + h)}{h} = 2x + h \). Now, take the limit as \( h \to 0 \): \( f'(x) = \lim_{h \to 0} (2x + h) = 2x \). Evaluate \( f'(3) = 2(3) = 6 \).