Use a linear approximation (or differentials) to estimate the given number. (Round your answer to five decimal places.)
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Consider the function f(x) = x ³. Since 2³ = 8 is a perfect cube that is pretty close to x = 1.999, take the linear approximation to f(x) at x = 2. It is

f(x) ≈ L(x) = f (2) + f ' (2) (x - 2)

Compute the derivative:

f(x) = x ³   →   f ' (x) = 3x ²   →   f ' (2) = 12

Then

L(x) = 8 + 12 (x - 2) = 12x - 16

Now approximate 1.999³ :

f (1.999) ≈ L (1.999) = 12•1.999 - 16 = 7.988

(Compare this to the actual value, 7.98801.)

Mix fraction: 3 1/27

Improper fraction: 82/27

Decimal approximation: 3.037

Step-by-step explanation:

What value is close to 28 that is a perfect cube...27 which equals 3^3.

So let's find the tangent line to the curve y=cubert(x) at x=27.

We will use this equation to approximate what happens at x=28.

First let's rewrite the radical in our equation;

y=x^(1/3)

Now differentiate

y'=(1/3) x^(1/3-1) by power rule

Simplify

y'=(1/3) x^(-2/3) or (1)/(3x^[2/3])

So the slope of our tangent line at x=27 is (1)/(3(27)^[2/3])=1/(3(3)^2)=1/(3×9)=1/27.

We will also need a point on this tangent lineWe know we have the point at x=27 because that is what our tangent line to curve is being found at.

So at x=27, we have y=cubert(27)=3. We used our equation y=cubert(x) here.

So we want to find the equation of the line that contains point (27,3) and has slope 1/27.

Point-slope form is

y-y1=m(x-x1)

Plug in our values

y-3=(1/27)(x-27)

Add 3 on both sides

y=3+(1/27)(x-27)

We will use this linear equation to approximate cubert(28) by replacing x with 28.

y=3+(1/27)(28-27)

y=3+(1/27)(1)

y=3+1/27

You can write that as a mix fraction if you want.

This value is than 3 but super close to 3 since 1/27 is close to 0.

Mix fraction: 3 1/27

Improper fraction: 82/27

Decimal approximation: 3.037

Cubert of 28 when smashed into calculator as is gives approximately 3.0366 which is pretty close to our approximation.

x + 4/7

step-by-step explanation:

use cymath