Question

Hi guys, can anyone help me with this, Thanks a lot:)

Answer 1

Answer:

  k = 1/9

Step-by-step explanation:

In order for the function to be continuous at x=9, the values of the two expressions must be the same at x=9.

The first expression evaluates to ...

  $\dfrac{\sqrt{9}-3}{-9}=-\dfrac{3-3}{9}=0$

The second expression needs to have the same value:

  $1 -k(9) = 0\\\\1 = 9k\\\\\boxed{k=\dfrac{1}{9}}$

Answer 2

Answer:

$k=1/9$

Step-by-step explanation:

A function is continuous at a point if and only if:

$\lim_{x \to n} f(x)=f(n)$

So, we have the piecewise function:

$f(x) = \left\{ \begin{array}{lI} \frac{\sqrt{x} -3}{-9} & \quad0< x <9 \\ 1-kx & \quad x\geq 9 \end{array} \right.$

And we want to find the value of k such that the function is continuous.

First, find the left hand limit of f(x):

$\lim_{x\to9^-} f(x)$

Since we're coming from the left, we'll use the first equation. Thus:

$=\lim_{x\to9^-} \frac{\sqrt{x}-3}{-9}$

Direct substitution:

$=\frac{\sqrt{9}-3}{-9}$

Simplify:

$=\frac{3-3}{-9}$

Subtract and divide:

$=\frac{0}{-9}=0$

So, what this tells us is that for the function to be continuous, the right hand limit as  f(x) approaches 9 from the right must also be equal to 0.

Therefore:

$\lim_{n \to 9^+} 1-kx=0$

Direct substitution:

$1-9k=0$

Subtract 1 from both sides:

$-9k=-1$

Divide both sides by -9:

$k=1/9$

Therefore, the value of k is 1/9.

So, our equation in the end is:

$f(x) = \left\{ \begin{array}{lI} \frac{\sqrt{x} -3}{-9} & \quad0< x <9 \\ 1-\frac{1}{9}x & \quad x\geq 9 \end{array} \right.$