Linear approximation

Question

Use linear approximation, i.e the tangent line, to approximate cube root 8.02 as follows. Let f(x)=cube root x and find the equation of the tangent line to f(x) at x=8 in the form y=mx+b.

m =

b=

using these values find the approximation cube root 8.02

Answer 1

Hi June,

If f(x)=x^1/3 then

df/dx=d(x^1/3)/dx=(x^1/3-1)/3=1/(3x^2/3)=1/{3[f(x)]²}

At x=8 ,

f(8)=8^1/3=2: df/dx=1/(3*f(8)²)=1/(3*4)=1/12

So, the equation of tangent line at the point (8,2) with the slope 1/12 would be as follows≈

y-2=(1/12)(x-8) or y=2+(1/12)(x-8)

Therefore for x=8.02

f(8.02)≈2+(1/12)*0.02≈2.00166

Now, how good is this result? To find out we'd need to find how would it change if the next (here, second) order of correction would be included

The 2nd order correction to our result would be the corresponding member of The Taylor series expansion near the point x=8

[d²f(x)/dx² ](x-8)^2/2

that could be found from

d²f(x)/dx²=d[df(x)/dx]/dx=d[1/[3f(x)²]=-[2/3f(x)³](df/dx)

which at x=8 is equal to

-[2/(3*2³)]/12=-(2/24)/12=-1/144

So, the 2nd order correction to the result 2.00166 of linear approximation would be

-(1/144)(0.02²/2)=0.00000139

So, our first order (linear, through tangent line) approximation was actually accurate up to the 5th decimal place (with the 2nd order correction it would become

(2.00166527667)

Let's compare that with the result returned by a calculator ; (8.02)^1/3=2.0016652797. We see that with 2nd correction included, the result becomes accurate up to the 8th decimal place

Hope this info (which might include some extra over what was required in the question) would be helpful.and give you a bit broader look at the essence of the approximate calculations

Best,

Dr.Ariel B.

Answer 2

Equation of the tangent line for f(x) at the point a:   y – f(a) = f’(a)(x-a)

In your question a = 8

f(8) = 2

f’(x)=1/(3 cube root (x²)) 

f’(8) = 1/ 12

Tangent line at x=8:  y – 2 = 1/12 (x - 8)

Simplified:   y = 1/12 x + 4/3

Now we use this linear approximation to find cube root 8.02 by plugging x=8.02 into it:

y = 1/12 * 8.02 + 4/3 = 2 and 1/600 or 1201/600