Question

use the definition of countinuity to find the value of k so that the function is continuous for all real numbers

Answer 1

First of all, recall the definition of absolute value:

$|x| = \begin{cases}x&\text{if }x\ge0\\-x&\text{if }x$

So if x < 4, then x - 4 < 0, so |x - 4| = -(x - 4), and the first case in h(x) reduces to

$\dfrac{|x-4|}{x-4}=\dfrac{-(x-4)}{x-4} = -1$

Next, in order for h(x) to be continuous at x = 4, the limits from either side of x = 4 must be equal and have the same value as h(x) at x = 4. From the given definition of h(x), we have

$h(4) = 5k-4\cdot4 = 5k-16$

Compute the one-sided limits:

• From the left:

$\displaystyle \lim_{x\to4^-}h(x) = \lim_{x\to4}\frac{|x-4|}{x-4} = \lim_{x\to4}(-1) = -1$

• From the right:

$\displaystyle \lim_{x\to4^+}h(x) = \lim_{x\to4}(5k-4x) = 5k-16$

If the limits are to be equal, then

-1 = 5k - 16

Solve for k :

-1 = 5k - 16

15 = 5k

k = 3