Question
Using the Implicit Function Theorem, show that there exists a unique differentiable function g in a neighbourhood of 1 such that in a neighbourhood of (,1,2) where
defines the function F. Also find g′( y).
Answer :
Word Count : 349
We are to use the Implicit Function Theorem to prove the existence of a unique differentiable function $g(y)$ such that $$ g(1) = 2 \quad \text{and} \quad F(g(y),y) = 0 $$ in a neighborhood of $(x_0,y_0) = (2,1)$, where $$ F(x,y) = x^5 + y^5 - 16xy^3 - 1. $$ Also, we need to compute $g'(y)$. --- ### Step 1: Verify $F(2,1)=0$ Check that $(x,y) = (2,1)$ satisfies $F(x,y)=0$: $$ F(2,1) = 2^5 + 1^5 - 16(2)(1)^3 - 1 $$ $$ F(2,1) = 32 + 1 - _________ _______ ________ ______ _____ _________ ___.
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We are to use the Implicit Function Theorem to prove the existence of a unique differentiable function $g(y)$ such that $$ g(1) = 2 \quad \text{and} \quad F(g(y),y) = 0 $$ in a neighborhood of $(x_0,y_0) = (2,1)$, where $$ F(x,y) = x^5 + y^5 - 16xy^3 - 1. $$ Also, we need to compute $g'(y)$. --- ### Step 1: Verify $F(2,1)=0$ Check that $(x,y) = (2,1)$ satisfies $F(x,y)=0$: $$ F(2,1) = 2^5 + 1^5 - 16(2)(1)^3 - 1 $$ $$ F(2,1) = 32 + 1 - _________ _______ ________ ______ _____ _________ ___.
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