contestada

Find the value of the constant k that makes the function continuous. ​g(x)equals=startset start 2 by 2 matrix 1st row 1st column startfraction 2 x squared minus 4 x minus 16 over x minus 4 endfraction 2nd column if x not equals 4 2nd row 1st column kx minus 16 2nd column if x equals 4 endmatrix 2x2−4x−16 x−4 if x≠4 kx−16 if x=4

Respuesta :

sqdancefan
You seem to have
.. g(x) = {(2x^2 -4x -16)/(x -4) . . . x ≠ 4
.. .. .. .. ..{ kx -16 . . . . . . . . . . . . . . x=4

The first expression can be simplified to
.. (2x^2 -4x -16)/(x -4) = 2(x +2)(x -4)/(x -4) = 2(x +2) . . . . x ≠ 4
At x=4, this simplified version has the value
.. 2(4 +2) = 12

To make the alternate definition of g(x) have that same value at x=4, we must have
.. k*4 -16 = 12
.. 4k = 28
,, k = 7

The constant k must be 7 for the function to be continuous at x=4.

Answer:

The value of k is 7.

Step-by-step explanation:

Given function,

[tex]f(x)=\left\{\begin{matrix}\frac{2x^2-4x-16}{x-4} & x\neq 4\\ kx-16 & x=4\end{matrix}\right.[/tex]

The function f(x) would be continuous, if,

[tex]\lim_{x\rightarrow 4} \frac{2x^2-4x-16}{x-4}=f(4)[/tex]

[tex]\lim_{x\rightarrow 4} 2(\frac{x^2-2x-8}{x-4})=4k-16[/tex]

[tex]\lim_{x\rightarrow 4} 2(\frac{x^2-4x+2x-16}{x-4})=4k-16[/tex]

[tex]\lim_{x\rightarrow 4} 2(\frac{x(x-4)+2(x-4)}{x-4})=4k-16[/tex]

[tex]\lim_{x\rightarrow 4} 2(\frac{(x+2)(x-4)}{x-4})=4k-16[/tex]

[tex]\lim_{x\rightarrow 4} 2(x+2)=4k-16[/tex]

[tex]2(4+2) = 4k-16[/tex]

[tex]2(6)=4k-16[/tex]

[tex]12+16 = 4k[/tex]

[tex]\implies 4k=28\implies k=7[/tex]

Hence, the value of k would be 7.

Otras preguntas