Let f and g be differentiable functions on [a, b]. Suppose |f'(x)|≥ g'(x), and g'(x) 0, Vx Є [a, b]. Prove: f(b) f(a)| ≥ |g(b) − g(a)|. -
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- Let h(x) = f(g(x)). If: f (3) = 0, ƒ'(42) = -5, f'(3) = 4, g(3) = 42, & g' (3) = 0, then h' (3) = Check, by definition not directLet f(x) be a function that is differentiable for all x. Let g(x) be defined by g(x) = f(x) + f(3 – x). Show that g'(x) has a root in the interval (0,3). (a) (b) Let f(x) be continuous and differentiable on the interval [-8,0]. Suppose f(-8) = -2 and f'(x) < 3 for all x. What is the largest possible value for f (0)? Justify your answer.
- Let f(x) = 11 – x², x > 0. Find and give the domain for f-(x). f-'(x) where a ?Let f be a differentiable function such that f(x + y) = f(x)+ f(y) + 5xy and f(h) lim h0 h = 16. Find f'(2).3. Let f (x) = (3x2 + 1)?. Find f'(x)in 3 different ways by following the instructions below in parts a, b and c: a) Algebraically multiply out the expression for f (x) and expand, then take the derivative. b) View f (x) as (3x? +1)(3x2 + 1) and use the product rule to find f' (x). C) Apply the chain rule directly to the expression f (x) = (3x² + 1)?. d) Are your answers in parts a, b, c the same? Why or why not?
- Suppose f: (a, b) → R is differentiable. True or false: If f is unbounded, then f' (the derivative of f) must be unbounded. True FalseLet f and g be functions satisfying that f(1) = 3, g(1) = 4, f'(1) = 1, g'(1) = 5. For the following function h(x), find h' (1). Answer: h(x) = Answer: Let f and g be functions satisfying that f(1) = 3, g(1) = 4, ƒ'(1) = 1, g'(1) = 5. For the following function h(x), find h' (1). f(x) g(x) h(x) = 2f(x) + 3g(x) f(x) + g(x)