Prove: If A is invertible, then is invertible and

Minimum and maximum values Let c be a number in the domain of f. f(c) is a local max if f(c) ≥ f(x) when x is near c. f(c) is a local min if f(c) ≤ f(x) when x is near c Fermat’s Theorem: If f has a local max or min at c, and if f’(c) exists, then f’(c)’=0 Be careful: The converse of this theorem is not always true. Consider f(x) = x^3 f’(x) = 3x^2 f’(0) = 3*0^2 =0 However there is no min/ max at x=0 the tangent line is horizontal there. Consider f(x) = [x] F has a minimum at x = 0; however f’(0) does not exist. Consider f(x) = √x f has a minimum at x = 0 f’(x) = 1/√x f’(0) = 1/2 0 d oes not exist The tangent line is vertical there Def. a critical number of a function f is a number c in the domain of f such that either f’(c) = 0 or f’(c) does not exist Theorem: If a function has a min or max it occurs at a critical number of f 3 Ex. find all the critical numbers of f(x) = x√-x3 Domain: all real numbers f(x) = x-3x^⅓ f’(x) = 1 - 3*1/3x^-⅔ = 1- 1/x^⅔ = 1 - 1/√x2 x = 0 is in the domain of f, but f’(0) does not exist f(x) = x - √x so 0 is a critical number now set f’(x) equal to zero and solve 1 - 1/√x2 = 0 √3x2 *1 = 1/√x * √ x2 3 ( x ) 3 = 1^3 x^2 = 1 x = -1, 1 So the critical numbers are -1, 0, 1 f(-1) = 2 this is a local max at x = 0 the tangent line is vertical but there is no max or min f(1) = -2 this is a local min Find the exact value of the minimum of f(x) = x^2ln(x) Domain x>0 f’(x) = x^2*(1/x)+2x ln(x) =x + 2x ln(x) Solve x + 2x ln(x) = 0 x