$\begin{array}{c}x^{2}-6 x+11=y \\ x=y+1\end{array}$

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Find higher derivatives of rational and radical functions

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\begin{array}{c}x^{2}-6 x+11=y \\ x=y+1\end{array}

Identify Equations: First, let's identify the system of equations we need to solve: $\newline$ ${x^{2}-6x+11=y},$ $\newline$ ${x=y+1}$ $\newline$ We need to find the value of $x$ that satisfies both equations.
Substitute and Solve: We can substitute the second equation into the first to solve for $x$ . Substitute $y$ from the second equation ( $x = y + 1$ ) into the first equation ( $x^2 - 6x + 11 = y$ ): $x^2 - 6x + 11 = x - 1$
Solve Quadratic Equation: Now, let's solve the resulting equation for $x$ : $x^2 - 6x + 11 = x - 1$ Move all terms to one side to set the equation to zero: $x^2 - 6x + 11 - x + 1 = 0$ $x^2 - 7x + 12 = 0$
Factor and Solve: Factor the quadratic equation: $\newline$ $(x - 3)(x - 4) = 0$
Check $x = 3$ : Set each factor equal to zero and solve for $x$ : $x - 3 = 0$ or $x - 4 = 0$ $x = 3$ or $x = 4$
Check $x = 4$ : We have two potential solutions for $x$ : $3$ and $4$ . We need to check which, if any, satisfy both original equations. $\newline$ First, check $x = 3$ : $\newline$ Substitute $x = 3$ into the second equation ( $x = y + 1$ ): $\newline$ $3 = y + 1$ $\newline$ $y = 2$ $\newline$ Now, substitute $x = 3$ and $y = 2$ into the first equation ( $x$ $1$ ): $\newline$ $x$ $2$ $\newline$ $x$ $3$ $\newline$ $x$ $4$ $\newline$ This is true, so $x = 3$ is a solution.
Check $x = 4$ : We have two potential solutions for $x$ : $3$ and $4$ . We need to check which, if any, satisfy both original equations. $\newline$ First, check $x = 3$ : $\newline$ Substitute $x = 3$ into the second equation ( $x = y + 1$ ): $\newline$ $3 = y + 1$ $\newline$ $y = 2$ $\newline$ Now, substitute $x = 3$ and $y = 2$ into the first equation ( $x$ $1$ ): $\newline$ $x$ $2$ $\newline$ $x$ $3$ $\newline$ $x$ $4$ $\newline$ This is true, so $x = 3$ is a solution.Next, check $x = 4$ : $\newline$ Substitute $x = 4$ into the second equation ( $x = y + 1$ ): $\newline$ $x$ $9$ $\newline$ $3$ $0$ $\newline$ Now, substitute $x = 4$ and $3$ $0$ into the first equation ( $x$ $1$ ): $\newline$ $3$ $4$ $\newline$ $3$ $5$ $\newline$ $3$ $6$ $\newline$ This is also true, so $x = 4$ is also a solution.