Question

Let f : \(\R^3 → \R\) be defined as f(x, y, z) = x³ + y³ + z³ , and let L : \(\R^3 → \R\) be the linear map satisfying
\(\lim\limits_{(x,y,z)\rightarrow(0,0,0)}\frac{f(1+x,1+y,1+z)-f(1,1,1)-L(x,y,z)}{\sqrt{x^2+y^2+z^2}}=0.\)
Then L(1, 2, 4) is equal to ____________. (rounded off to two decimal places)

Answer 1

Correct Answer: 20.99 - 21.01

To solve the problem, we need to determine the linear map \( L: \mathbb{R}^3 \to \mathbb{R} \) such that:

\(\lim_{(x,y,z)\rightarrow(0,0,0)}\frac{f(1+x,1+y,1+z)-f(1,1,1)-L(x,y,z)}{\sqrt{x^2+y^2+z^2}}=0.\) 

First, evaluate \( f(1,1,1) \):

As given, \( f(x,y,z)=x^3+y^3+z^3 \), so:

\(f(1,1,1)=1^3+1^3+1^3=3.\)

Now calculate \( f(1+x,1+y,1+z) \):

\((1+x)^3+(1+y)^3+(1+z)^3 = 1+3x+3x^2+x^3 + 1+3y+3y^2+y^3 + 1+3z+3z^2+z^3. \)

This simplifies using linear terms:

\( = 3 + 3x + 3y + 3z + \text{(higher-order terms)}. \)

Subtracting \( f(1,1,1) \), we have:

\( f(1+x,1+y,1+z)-f(1,1,1) = 3x + 3y + 3z + \text{(higher-order terms)}. \)

By the definition of the linear map \( L(x,y,z) \), drop higher-order terms. Therefore:

\(L(x,y,z) = 3x + 3y + 3z.\)

Next, find \( L(1,2,4) \):

Substitute \( x=1 \), \( y=2 \), \( z=4 \) into \( L(x,y,z) = 3x + 3y + 3z \):

\(L(1,2,4) = 3(1) + 3(2) + 3(4) = 3 + 6 + 12 = 21.\)

Finally, check if \( 21 \) falls within the expected range, which is \( (20.99,21.01) \).

Yes, \( 21 \) is within the range.