course Mth 271 LmԞpǷassignment #022
......!!!!!!!!...................................
21:51:35 Query 3.1.4 derivative of -3x `sqrt(x+1) at x=-1, -2/3, 2`sqrt(3)/3 **** What is the derivative of the function at each of the indicated points?
......!!!!!!!!...................................
RESPONSE --> I first got the derivative then filled in for the x values f(x)=-3x(x+1)^1/2 f'(x)=-3x(x+1)^1/2 + (x+1)^1/2 (-3x) f'(x)=-3x(2(x+1)^1/2)(-3x) f'(x)=-3x(2x+1)(-3x) f'(x)= -6x^2 + 3x(-3x) f'(x)= -6x^2 + 9x^2 f'(x)= -12x + 18x Filling in for -2/3 -12(-2/3) + 18(-2/3) 8 + -12 = -4 Filling in for -1 -12(-1) + 18(-1) 12 + -18 = -6 Filling in for 0 -12(0) + 18(0) = 0 confidence assessment: 2
.................................................
......!!!!!!!!...................................
21:56:02 f(x) = -3x(x + 1)^(1/2) . Using the Product Rule we get f ' (x) = (-3) ( x + 1)^(1/2) + 1/2(x +1)^(-1/2)(1)(-3x) = -3(x+1)^(1/2) - 3/2 * x (x+1)^(-1/2). You could substitute into this form and get the right answers. However it's good to see how to simplify this expression: Simplifying, we multiply the first term by (x+1)^(1/2) / (x+1)^(1/2) to obtain f ' (x) = -3(x+1) / (x+1)^(1/2) - (3 x / 2)/(x+1)^(1/2) = [-3(x+1) - 3 x / 2 ] / (x+1)^(1/2) = (-3x - 3 - 3 x / 2) / (x+1)^(1/2) = (-6x - 6 - 3x) / ( 2(x+1)^(-1/2) ) = -3 ( 3x + 2) / ( 2(x+1)^(-1/2) ) Thus f ' (x) = -3(3x + 2) / (2(x +1)^(1/2)) f ' (-1) is undefined (denominator is zero) f ' (0) = -6/2 = -3 f ' (-2/3) = 0 (critical number, numerator is zero) DER
......!!!!!!!!...................................
RESPONSE --> I used the product rule but I didn't get the same answer. I have made a note of how you got this answer. self critique assessment: 2
.................................................
......!!!!!!!!...................................
21:59:26 Query 3.1.8 increasing, decreasing behavior of x^2 / (x+1) On what open intervals is x^2 / (x+1) increasing, and on what intervals is it decreasing?
......!!!!!!!!...................................
RESPONSE --> f(x)=x^2/x+1 (x+1)(2x) - x^2(x+1)/(x+1)^2 2x + 2x^2 - x^2/(x+1)^2 = 2x + x^2/(x+1)^2 x(x+2)/(x+1)^2 = 0 f is increasing at (- infinity, -2) and (0, infinity) decreasing at (-1,0) and (-2,0) confidence assessment: 2
.................................................
......!!!!!!!!...................................
22:02:14 We first look at where the function itself is positive, where it is negative, and where it is undefined: The numerator of the function f(x) is always positive, being a square. The denominator of f(x) goes from positive to negative at x = 1 and stays there, so the function is negative on the first two intervals, positive on the other two. To determine where the function is increasing and where decreasing we need to find the derivative and determine where it is positive and where it is negative. Using the quotient rule we get f ' (x) = (2x( x + 1) - (1)(x^2)) / (x + 1)^2 f ' (x) = (2x^2 + 2x - x^2) / (x + 1)^2 f ' (x) = (x^2 + 2x) / (x + 1)^2 f ' (x) = (x(x + 2)) / (x +1)^2 0 = (x(x + 2)) / (x + 1)^2 . The derivative is zero at x = 0 and x = -2, and undefined at x = -1. So the function can change sign only at -2, -1 or 0. The function therefore has the same sign on each of the intervals (-infinity, -2), (-2, -1), (-1, 0), (0 infinity). The derivative goes from positive to negative at -2 and from negative to positive at 0. So the function is decreasing on (-2, 0) and increasing otherwise. So the function is negative and increasing on (-infinity, -2), then negative and decreasing on (-2, -1), where it decreases asymptotic to the line x = 1. Then the function is positive and decreasing on (-1,0), and positive and increasing on (0, infinity). Note that x = -2 and x = 0 are extrema, with the function reaching a relative maximum of -4 at x = -2 and a relative minimum of 0 at x = 0. **
......!!!!!!!!...................................
RESPONSE --> I got most of this correct. I didn't give as much information as you did but I understand how you got the answers. self critique assessment: 2
.................................................
......!!!!!!!!...................................
22:04:04 ** Query 3.1.30 critical numbers, increasing and decreasing intervals for f(x) = x / (x+1) What are the critical numbers for the given function, and on what intervals is it increasing, and on what intervals is it decreasing?
......!!!!!!!!...................................
RESPONSE --> f'(x)= x+1(1) - x(-1)/(x+1)^2 = 1/(x+1)^2 Discontinuous:x = -1 Critical number: -1 it is increasing at (- infinity, 1) and (1, infinity) self critique assessment: 2
.................................................
......!!!!!!!!...................................
22:05:10 Critical numbers are the numbers where the derivative is zero, and the points where the function is discontinuous. f( x ) = x / (x +1) . Using the quotient rule we get: f ' (x) = (1(x + 1) - (1)(x)) / (x + 1)^2 f ' (x) = 1 / (x + 1)^2 f ' (x) = 1 / (x + 1)^2. We can set this expression equal to zero: 1 / (x+1)^2 = 0. However we will get no solutions, since the numerator is never zero. The only possible critical numbers are those where the function is discontinous. This occurs only at x = 1, which makes the denominator zero. The function is discontinuous at x = 1. So x = 1 is a critical number. The derivative, being the reciprocal of a perfect square, is positive at numbers less than 1 and is positive at numbers greater than 1. The derivative in undefined, and approaches infinity, as x -> -1. On the interval (-infinity, 1), the derivative is positive and the function is therefore increasing. On the interval (1, infinity), the derivative is positive and the function is therefore again increasing. **
......!!!!!!!!...................................
RESPONSE --> I got this one correct. self critique assessment: 2
.................................................
"