10. given that the function\n$f(x)=\\begin{cases}3x + 2, & x lt k \\\\ 2x - 3, & k \\leq x \\leq 8…

10. given that the function\n$f(x)=\\begin{cases}3x + 2, & x lt k \\\\ 2x - 3, & k \\leq x \\leq 8 \\end{cases}$\nis continuous on (-inf,8, find k.

10. given that the function\n$f(x)=\\begin{cases}3x + 2, & x lt k \\\\ 2x - 3, & k \\leq x \\leq 8 \\end{cases}$\nis continuous on (-inf,8, find k.

Answer

Explanation:

Step1: Use the definition of continuity

For a function ( f(x) ) to be continuous at ( x = k ), ( \lim_{x\rightarrow k^{-}}f(x)=\lim_{x\rightarrow k^{+}}f(x)=f(k) ). Since ( \lim_{x\rightarrow k^{-}}f(x)=3k + 2 ) (using ( y=3x + 2 ) for ( x<k )) and ( \lim_{x\rightarrow k^{+}}f(x)=2k-3 ) (using ( y = 2x-3 ) for ( k\leq x\leq8 )) and ( f(k)=2k - 3 ). Set ( 3k+2=2k - 3 ).

Step2: Solve the equation for ( k )

Subtract ( 2k ) from both sides of the equation ( 3k+2=2k - 3 ). We get ( 3k-2k+2=2k-2k - 3 ), which simplifies to ( k+2=-3 ). Then subtract 2 from both sides: ( k=-3 - 2 ).

Answer:

( k=-5 )