Question

Use differentials to find an approximate value of the firth root of 29

Answer 1

Hint: What's the integer 5th power closest to 29?

Answer 2

looks like 32 is the closest integer fifth power to 29

Answer 3

so suppose we have
f(x) = x^(1/5)
and we want to approximate f(29). we are going to make a linear approximation using the derivative

Answer 4

Right. Thanks.
So, what is the function in question here? f(x) = ???? Note: f(32) would be 2.

Answer 5

lets make a linear approximation at x = 32 for f(x) = x^1/5
so using point slope equation of line we have:
y - f(32) = f ' (32) ( x - 32)

Answer 6

@perl, it's fine if you want to take over this conversation, but if you do, please lead @mojalefa towards finding his or her own solution. You are currently doing all the work.

Answer 7

You might want to introduce the concept of "linear approx." BEFORE actually using it.

Answer 8

f ' (x) = 1/5 * x ^(-4/5)
f ' (32) = 1/5 * 32 ^(-4/5)= 1/80

Answer 9

there is more than one way to answer this question

Answer 10

so the linear approximation is :
y - 2 = 1/80 ( x - 32)
y = 1/80*x - 32/80 + 2

Answer 11

now test to see if this equation is correct.
plug in x = 32
y(32) = 1/80 *32 - 32/80 + 2 = 2

Answer 12

Not if you're going to use linear approximations. Differentials are closely related to linear approximations and vice versa. My point: Please guide mojalefa towards finding his /her own solution. Explaining your method would also help.

Answer 13

ok now plug in x = 29

Answer 14

@perl: Have you read what I've typed above? Guide the other person towards finding his/her own solution.

Answer 15

y(29) = 1/80*29 - 32/80 + 2 = 1.9625

Answer 16

i am guiding, please see the steps above

Answer 17

now compare y(29) with the exact answer 29^(1/5)

Answer 18

thanks a million

Answer 19

please explain point slope equation

Answer 20

@mojalefa: would you please post each new question separately.

The point-slope equation is one form of the equation of a straight line, and is used when you are given at least one point on a line and the slope of that line, to find the equation of the line:\[y-y _{1}=m(x-x _{1})\]Have you seen and/or used this before?

Suppose you have a line with slope 2 and passing thru the point (2,3). What's the equation of that line?

Answer 21

Its my first time to see it.

Answer 22

y=2x+1

Answer 23

Please start with the standard point-slope form:\[y-y _{1}=m(x-x _{1})\] and then replace m with the given slope (2), and replace x1 with 2 and y1 with 3.

Please show me what you get. Then, in the end, simplify the equation. I'd like to see how you got your y=2x+1.

Answer 24

y-y1=m(x-x1) i made y the subject of the formula i.e y=m(x-x1)+y1 then substituted the value of m,x1 and y1 then simplified.is that correct?

Answer 25

As explained, the slope, m, is 2 and one point on the line in question is (2,3). Here's what we'd get by substitution:\[y-3 = 2(x-2)\]would you please simplify this?