Find the linear approximation of g(x) = 81+x8 at a=0. Use the linear approximation to approximate the value of 8√95 and 8√1.1

y=?

I read the function as ⁸√(1+x⁸)=(1+x⁸)^⅛.

The derivative is dg/dx=((1+x⁸)^-⅞)/8. When x=a=0, this is ⅛, and is the gradient for the linear equation. g(a)=g(0)=1.

We can write this linear equation as L(x)=x/8+c, where c is a constant found be plugging in (0,1), where L(0)=g(0)=1. So c=1.

y=L(x)=x/8+1 is an approximation to g(x). We can also replace x by x⁸ so:

y=L(x⁸)=x⁸/8+1.

(We can also write ∆g=∆x((1+x⁸)^-⅞)/8, where ∆x is a small change in x which creates a small change ∆g in g.)

To find ⁸√1.1=⁸√(1+0.1) we can use the linear approximation:

y=0.1/8+1=1.0125 or 1.01 to 2 decimal places.

To find ⁸√95 we need to reduce 95 to p⁸(1+q/p⁸) (p and q are constants) so that we have p(⁸√(1+q/p⁸) and y=q/(8p⁸)+1 as the approximation. Then we multiply this by p to get py as the final approximation solution. If p=2 and q=-161, we have 256(1-161/256)=95, so y=-161/2048+1=0.9214 approx.

And the approximation is 2×0.9214=1.84 to 2 decimal places.

by Top Rated User (796k points)