Charlie scale 9

Jake was asked to find whether

g (x) = x + 4 \ln (| x - 1 |)

has a relative minimum. This is his solution:

Step 1:

g^{'} (x) = \frac{x + 3}{x - 1}

Step 2: The critical values are

x = - 3

and

x = 1

Step 3:

Interval	Test $x$ ‍-value	$g^{'} (x)$ ‍	Verdict
$(- \infty, - 3)$ ‍	$x = - 4$ ‍	$g^{'} (- 4) = 0.2 > 0$ ‍	$g$ ‍ is increasing $↗$ ‍
$(- 3, 1)$ ‍	$x = 0$ ‍	$g^{'} (0) = - 3 < 0$ ‍	$g$ ‍ is decreasing $↘$ ‍
$(1, \infty)$ ‍	$x = 2$ ‍	$g^{'} (2) = 5 > 0$ ‍	$g$ ‍ is increasing $↗$ ‍

Step 4:

g

decreases before

x = 1

and increases after it, so

g

has a minimum point at

x = 1

Is Jake's work correct? If not, what's his mistake?

(Choice A)
Jake's work is correct.
(Choice B)
Step 1 is incorrect. Jake didn't differentiate $g$ ‍ correctly.
(Choice C)
Step 2 is incorrect. $g^{'} (1)$ ‍ isn't equal to zero so $x = 1$ ‍ isn't a critical value.
(Choice D)
Step 4 is incorrect. Jake didn't check whether $x = 1$ ‍ is in the domain of $g$ ‍.

US Math sandbox