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1. Average rate of change over specified interval.
f(x)= 4-4x-2x^2 over (3,4). I got -70. Does that seem correct?

2.Find the derivative. f(x)= 9x^2- 2x+7 . My answer is 14+18x. Correct or not?

3. Can someone tell me how to set up for determining the equation of tangent line at given point.

f(x)= (x-4)^2; (5,1)

Thanks.

2007-07-03 13:40:49 · 5 answers · asked by Anonymous in Science & Mathematics Mathematics

5 answers

If f(x) = 4 - 4x - 2x^2, then the slope is the first derivative

f'(x) = 0 - 4 -4x = -4x - 4.

but at x=3 slope is -12-4 = -16.
at x=4 slope is -16-4 = -20.

Since the slope changes linearly over the range, the average slope (rate of change) is -18.

More correctly, without assuming the slope changes linearly, to find the average slope for a function you would have to integrate the function of the slope and find the total area under the curve, divided by the range to get an arithmetic average of the slope (or rate of change) in that interval:

avg. slope = Int(x from 3 to 4) [-4x-4] dx / Range

(but Range = 4-3 = 1)

= (x from 3 to 4) [-2x^2 -4x] /1.

= (x=4)[-2x^2-4x] - (x=3)[-2x^2-4x]

= -32 - 16 + 18 +12 = -48 + 30 = -18 again.


2) If f(x) = 9x^2 -2x + 7,

f'(x) = 18x-2.

Remember that if f(x) = ax^n then f'(x) = anx^(n-1).

So take the rate of change of each term and add them up.

Rate of change of 9x^2 is 9*2*x = 18x.
Rate of change of -2x = -2(1)x^0 = -2(1)1 = -2.
Rate of change of +7 or 7x^0 = 7(0)x^(-1) = 0. [In other words, a constant is like a straight line y=7, since it never changes with x, it is flat, slope is 0. So constants never contribute to the rate of change or the slope.]

3) You can find the tangent line if you can find the slope (rate of change) of that function at that point, and that becomes your slope m in the general linear formula y = mx + b. Substitute the known (x,y) on this line (your tangent point from the function) and, knowing m, you can then find b.

f(x) = (x-4)^2 = (x-4)(x-4) = x^2 -4x -4x + 16 = x^2 - 8x +16.

So f'(x) = 2x -8.

At (5,1) the rate of change is 2(5)-8 = 2.

But that means the tangent at that point has slope 2.

y = mx+b = 2x + b.

Since you know that (5,1) is on this line where the tangent meets it, then

1 = 2(5) + b or b = 1-10 = -9.

So the correct tangent line is y = 2x -9.

2007-07-03 14:08:23 · answer #1 · answered by PIERRE S 4 · 1 1

1. = f(4) -f(3))/(4-3) = f(4) -f(3)
f(4) = 4 -16 -32 = -44
f(3) = 4-12 -18 = -26
f(4)-f(3) = -44-(-26) = -18

9x^2- 2x+7
derivative = 18x -2

Find the derivative which is the slope of the tangent.
slope = d/dx(x^2-8x+16) = 2x -8
So slope when x = 5 is 2*5 -8 = 2
So y= 2x + b
1=2*5 +b --> b = -9
So the tangent line is y=2x-9

2007-07-03 13:55:14 · answer #2 · answered by ironduke8159 7 · 1 0

1. Average rate of change = [f(x2) - f(x1)]/(x2 - x1)
f(3) = 4 - 12 - 18 = -26
f(4) = 4 - 16 - 32 = -44
Average rate of change = (-44 - (-26)/(4-3)
= -18

2. f(x) = 9x^2 - 2x + 7
f'(x) = 18x - 2
Remember the power rule: f'(x^n) = nx^(n-1)

3. The slope of a curve is its derivative
f(x) = (x-4)^2
f'(x) = 2(x-4) = 2x-8
x=5
f'(5) = 10-8 = 2
Equation of a straight line is y=mx+c
m=2, x=5, y=1
1 = 5*2 + c
c = -9
y = 2x - 9
2x - y = 9

2007-07-03 13:48:29 · answer #3 · answered by gudspeling 7 · 1 0

1) Average rate of change
= [f(4) - f(3) ] / (4-3)
= -44 - (-26) = -18

You can test the answer by evaluating f'(3.5)
f'(x) = -4 - 4x
f'(3.5) = -18 exactly

2) It's 18x - 2

3) f'(x) = 2*(x-4)
f'(5) = 2 = slope of line at x = 5
eqn: y - 1 = 2*(x - 5)
y = 2x - 9

2007-07-03 13:45:57 · answer #4 · answered by Dr D 7 · 2 0

1. the answer is -18
2. y'=18x-2
3. find the derivative; solve for m=? or the slope so sub in y
then put it in the point slope formula
Hope that helped! good luck

2007-07-03 14:03:16 · answer #5 · answered by Luna_5 2 · 0 0

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