University:
York University
Course:
MATH 1013 | Applied Calculus I
Academic year:

2022
Views:

218

Pages:

4
Author:

Nick T.

**Homework 27.4.28** **Definition 1:** m = lim(h→0) [f(a+h) - f(a)] / h **Definition 2:** m = lim(x→a) [f(x) - f(a)] / (x - a) **Problem 1:** f(x) = 6x - x² **a)** f'(a) = lim(h→0) [f(a+h) - f(a)] / h = lim(h→0) [6(a+h) - (a+h)² - (6a - a²)] / h = lim(h→0) [6a + 6h - a² - 2ah - h² - 6a + a²] / h = lim(h→0) [4h - h²] / h = lim(h→0) 4 - h = 4 **b)** f'(a) = lim(x→a) [f(x) - f(a)] / (x - a) = lim(x→a) [(6x - x²) - (6a - a²)] / (x - a) = lim(x→a) [6x - x² - 6a + a²] / (x - a) = lim(x→a) [(x - a)(6 - x - a)] / (x - a) = lim(x→a) 6 - x - a = 6 - a - a = 6 - 2a **Tangent line slope:** f'(1) = 4 **Tangent line equation:** y - f(a) = f'(a)(x - a) y - 5 = 4(x - 1) y = 4x + 1 **Problem 2:** f(x) = √x **a)** f'(a) = lim(h→0) [f(a+h) - f(a)] / h = lim(h→0) [√(a+h) - √a] / h = lim(h→0) [(√(a+h) - √a) * (√(a+h) + √a)] / (h * (√(a+h) + √a)) = lim(h→0) [(a+h) - a] / (h * (√(a+h) + √a)) = lim(h→0) 1 / (√(a+h) + √a) = 1 / (2√a) **b)** f'(a) = lim(x→a) [f(x) - f(a)] / (x - a) = lim(x→a) [√x - √a] / (x - a) = lim(x→a) [(√x - √a) * (√x + √a)] / ((x - a) * (√x + √a)) = lim(x→a) (x - a) / ((x - a) * (√x + √a)) = lim(x→a) 1 / (√x + √a) = 1 / (2√a) **Tangent line slope:** f'(1) = 1/2 **Tangent line equation:** y - f(a) = f'(a)(x - a) y - 1 = 1/2(x - 1) y = 1/2x + 1/2