Find the Difference Quotient for f(x)=2x^2-3x+2?
With f(x) = 2x^2 - 3x + 2, we have: f(x + h) = 2(x + h)^2 - 3(x + h) + 2. Then, the difference quotient is: [f(x + h) - f(x)]/h = [2(x + h)^2 - 3(x + h) + 2 - (2x^2 - 3x + 2)]/h = (2x^2 + 4xh + 2h^2 - 3x - 3h + 2 - 2x^2 + 3x - 2)/h = (4xh + 2h^2 - 3h)/h, by simplifying = [h(4x + 2h - 3)]/h, by factoring out h = 4x + 2h - 3, by canceling h. I hope this helps!
How do I find the Difference Quotient of 1/sqrt(x)?
f(x) = 1/√x Difference quotient = (f(x+h) − f(x)) / h = (1/√(x+h) − 1/√x) / h = √x √(x+h) (1/√(x+h) − 1/√x) / (h √x √(x+h)) = (√x − √(x+h)) / (h √x √(x+h)) To get rid of h in denominator, we can multiply numerator and denominator by (√x + √(x+h)) = (√x − √(x+h)) (√x + √(x+h)) / (h √x √(x+h) (√x + √(x+h))) = (x − (x+h)) / (h √x √(x+h) (√x + √(x+h))) = (−h) / (h √x √(x+h) (√x + √(x+h))) = −1 / (√x √(x+h) (√x + √(x+h))) Now the reason we put difference quotient in this form is because we can now find slope as h approaches 0 (i.e. slope at point (x, 1/√x)), which is −1 / (√x √x (√x + √x)) = −1 / (x (2√x)) = −1/(2x√x) = −1/(2x^(3/2))
Find the difference quotient. f(x) = 7x^2 - 3x + 2?
f (x + h) = 7 * (x + h)² - 3 * (x + h) + 2 <=> f (x + h) = 7x² + 14xh + h² - 3x - 3h + 2 <=> f (x + h) - f (x) = 7x² + 14xh + h² - 3x - 3h + 2 - 7x² + 3x - 2 = 14xh - 3h + h² <=> (f (x + h) - f (x)) / h = 14x - 3 + h => lim (f (x + h) - f (x)) / h = 14x - 3 h -> 0 Conclusion : The derivation of f (x) is 14x - 3
Is there a way to find the difference quotient using a TI-83 Plus calculator?
The only function I know of on the TI-83 is to calculate the value of the derivative of a function. Example: what is the value of the derivative of f(x) = 3x^3 - 42x at x = 3. Press the Math key Arrow down to 8:nDeriv enter input:(3x^3-42x,x,3) enter answer = 39. For your limit definition method above... lim h-->0 [3(x+h)^3-42(x+h) - 3x^3+42x]/h = lim h-->0 [3x^3+9x^2h+9xh^2+3h^3-42x-42h-3x^2+42x]... = lim h-->0 [9x^2h+9xh^2+3h^3-42h]/h = lim h-->0 [9x^2 + 9xh + 3h^2 - 42] taking the limit h-->0 = 9x^2 - 42
(a) Find the value of the difference quotient for s(x) = xx at x = 3 for each value of h below. Give your answers to 4 decimal places.?
f(x) = x^x h= 0.1; (f(3.1)- f(3))/h = ((3.1)^(3.1) - (3)^3 ) / 0.1 = 63.5963 h=0.01 ; (f(3.01)-f(3))/0.01 = ((3.01)^(3.01) - (3)^3 ) / 0.01 = 57.307180 h=0.001 ; (f(3.001)-f(3))/0.001 = ((3.001)^(3.001) - (3)^3 ) / 0.001 = 56.7285387 h=0.0001 ; (f(3.0001)-f(3))/0.0001 = ((3.0001)^(3.0001) - (3)^3 ) / 0.0001 = 56.6689279 h=0.00001 ; (f(3.00001)-f(3))/0.00001 = ((3.00001)^(3.00001) - (3)^3 ) / 0.00001 = 56.6631714 s'(3) = 56.66
Given 5x^2 - 4x find the difference quotient?
[f(x+h)-f(x)]/h f(x+h)-f(x)=5(x+h)^2-4(x+h)-f(x)= 5x^2+10xh+5h^2-4x-4h-5x^2+4x= 10xh+5h^2-4h. (10xh+5h^2-4h)/h = 10x+5h-4 The difference quotient is: 10x+5h-4
Find the difference quotient for the function.?
So you have the formula : [f(x + h) - f(x)] / h. 1. f(x) = -4x - 1 To convert f(x) into f(x + h), you just replace x with x + h. so wherever you see x in the equation, replace it with x + h. Thus, f(x + h) = -4(x + h) - 1 Now we can substitute these into the difference quotient. [f(x + h) - f(x)] / h = {[-4(x + h) - 1] - [-4x - 1]} / h = (-4x - 4h - 1 + 4x + 1) / h = -4h / h = -4 2. f(x) = (-1/3)(x - 5), so, f(x + h) = (-1/3)(x + h - 5) DQ = [(-1/3)(x + h - 5) - (-1/3)(x - 5)] / h = (-x/3 - h/3 + 5/3 + x/3 - 5/3) / h = (-h/3) / h = -1/3 3. f(x) = x^2/4, so, f(x + h) = (x + h)^2/4 DQ = [(x + h)^2/4 - x^2/4] / h = (x^2/4 + xh/2 + h^2/4 - x^2/4) / h = (xh/2 + h^2/4) / h = x/2 + h/4
How do you find the difference quotient for this given function. f(x)= 4+3x-x^2, f(3+h) - f(3)/h?
just plug in (3+h) for x, then 3 for x 4 + 3(3+h) - (3+h)^2 - (4 + 3(3) - 3^2) all over h then do it out: 4 + 9 + 3h - (9 + 6h + h^2) - (4+9-9) over h then simplify: 13 + 3h - 9 - 6h - h^2 - 4 over h = -3h - h^2 over h then factor out an h: h(-3 - h) over h and cancel the h leaving -3 - h which, as h → 0 is -3