How do you find the average rate of change of [math]f (x) = x^2 - 4x + 1[/math] from [math]-4[/math] to [math]-3[/math]?
You could first find the rate of change as a function of [math]x[/math]. That's just the derivative: [math]f'(x) = 2x-4[/math]. You could then find the average value of that function on the interval (-4, -3). That's just the integral over the interval of the function divided by the length of the interval: [math]\frac 1{-3 -(-4)} \int_{-4}^{-3} (2x-4) dx [/math] [math]= (x^2-4x)|_{x=-4}^{-3}=-11 [/math].But that would be a silly approach. You'd be using Calculus without understanding Calculus. (Which is common enough, but not something to strive for...) Because of course, the average rate of change of a function is the total change divided by the length of the interval. So the answer must be: [math]\frac{f(-3)-f(-4)}{-3-(-4)}[/math] [math] =f(-3)-f(-4) = 22-33=-11 [/math].
Find the average rate of change of f(x)=-2-4x^-1?
Same as i said before, but now the derivative of f (x) will be 4x^-2. And this is your average rate of change. (x^m)' = mx^m-1 here is another method with calculus using the definition of average rate of change: lim ((-2-4(x+h)^-1) - (-2-4x^-1)) / h h ->0 = (-2-(4/x+h) + 2 + 4/x) /h = (-2 -(4/x+h) +2 + 4/x) /h = (-4/(x+h) + 4/x ) / h = (-4x/x(x+h) + 4(x+h)/x(x+h)) / h -using common denominator = ((-4x + 4(x+h)/ x(x+h)) /h = (4h)/(x(x+h))/h =4/ x(x+h) as h approaches zero.... = 4 / x^2 or 4x^-2
How do I find the average rate of change?
You have the correct definition of average rate of change. I think you might have made a small calculation mistake.My guess is that you read f(x) wrong, did you interpret f(x) = x - 1 or f(x) = x^(2–1) when doing this problem? (f(b) - f(a)) / (b - a) = 1 in both these cases.Also, it could be a problem with your calculator if you’re using one to solve this question, so in that case, try resetting your calculator, or if that doesn’t work, try to be more careful with your order of operations.
Find the average rate of change for the given interval. R theta= square root of 3 theta + 1 [0,5]?
R(theta) = sqrt(3 theta + 1) , interval: (0, 5) AROC R(theta) = [R( 5) - R(0) ]/ (5 - 0) AROC R(heta) = [ sqrt(3*5 + 1) - sqrt(3*0 + 1] / 5 AROC R(theta) = [ sqrt(16) - sqrt(1)] / 5 AROC R(theta) =( 4 - 1) / 5 AROC R(theta) = 3/5 .....<===Aansswerr
What is the rate of change between the interval x = 1 and x = three pi over two?
This is an incomplete question. What function or scenario ard you looking to compute a rate of change for? Just giving an interval does no good.
Calc-Find the average rate of change of the function over the given interval(s):R(x) = squareroot(4x+1); [0,2]
Not sure whether I interpreted the question properly. When x=0, sqrt(4x+1) = Sqrt(1) =+/- 1 When x = 2, Sqrt(4x+1) =Sqrt(8+1) = Sqrt(9) = +/- 3 See, there are two curves here. Now the change in x is 2 - 0 =2 The change is R(x) is: (there are two possibilities:) 3 - 1 = 2 and -3 - -1 = -2 The average rate of change is (change in y) / (change in x) = (+/- 2 ) / 2 = +/- 1 So there are two possible answers: + 1 for one curve, and - 1 for the other.
Limits: Find the average rate of change of f(x) = (square root of 4x+1) over the interval [0,2]?
The average rate of change is Δy / Δx. That's the change in y (rise) divided by the change in x (run). You find the two coordinates for x=0 and x=2. f(0) = √(4×0+1) = 1 f(2) = √(4×2+1) = 3 The change in y is 3-1=2. The change in x is 2-0=2. The average rate of change is: 2/2 = 1
Find the average rate of change for the function?
The average rate of change over an interval is the same as the slope between two points. So, let's find those two points first: x = 2 y = 2^2 + 5(2) y = 4 + 10 = 14 Point (2, 14) x = 5 y = 5^2 + 5(5) y = 25 + 25 = 50 Point (5, 50) To find a slope when given two points, the formula is: m = (y2 - y1) / (x2 - x1) So, let's use (2, 14) as (x1, x2) and (5, 50) as (y1 - y2): m = (50 - 14) / (5 - 2) m = (36) / (3) m = 12 And 12, folks, is your average rate of change. [cue applause]
Average rate of change of square root of x between x=4 and x=9?
The average rate of change of f(x) from x=a to x=b is: [f(b) - f(a)] / (b - a). We see that: f(9) = √9 = 3 f(4) = √4 = 2. Then, the average rate of change of f(x) = √x from x=4 to x=9 is: [f(9) - f(4)] / (9 - 4) = (3 - 2) / 5 = 1/5. <== ANSWER I hope this helps!
Find the average rate of change of the function: ?
f(x)=SQRT(5x + 6) f(2)= SQRT(16)=4 f(0) = SQRT(6) Average rate of change = [f(2)-f(0)] / (2-0) = [ 4 - √6] / 2 = 2 - √6 / 2