Question

the graph of a function $f(x)$ is shown below:
for how many values of $x$ is $f(x)$ not differentiable on the interval $(-2, 5)$?
$\bigcirc$ 2
$\bigcirc$ 5
$\bigcirc$ 3
$\bigcirc$ 1
$\bigcirc$ 4

Explanation:

Step1: Recall non - differentiability conditions

A function is not differentiable at a point if:
1. The function is not continuous at that point (jump discontinuity, removable discontinuity, infinite discontinuity).
2. The function has a corner (sharp turn) at that point.
3. The function has a vertical tangent at that point.

Step2: Analyze the graph

- **Point \(x = - 1\)**: There is a removable discontinuity (open circle), so the function is not continuous here, hence not differentiable.
- **Point \(x = 1\)**: There is a vertical asymptote (infinite discontinuity), so the function is not continuous (and has an infinite discontinuity) here, hence not differentiable.
- **Point \(x = 2\)**: There is a jump discontinuity (open circle and closed dot), so the function is not continuous here, hence not differentiable.
- **Point \(x = 4\)**: There is a sharp corner (the graph changes direction abruptly), so the function has a corner here, hence not differentiable.

Wait, let's re - evaluate. Wait, the interval is \((-2,5)\). Let's list the points:

1. At \(x=-1\): Removable discontinuity (open circle), not differentiable.
2. At \(x = 1\): Vertical asymptote (infinite discontinuity), not differentiable.
3. At \(x = 2\): Jump discontinuity (open circle and closed dot), not differentiable.
4. At \(x = 4\): Sharp corner (change in slope), not differentiable. Wait, but maybe I made a mistake. Wait, let's check again.

Wait, the conditions for non - differentiability:
- Discontinuities (removable, jump, infinite): At \(x=-1\) (removable), \(x = 1\) (infinite), \(x = 2\) (jump).
- Corners or cusps: At \(x = 4\) (corner). Also, is there another point? Wait, maybe I miscounted. Wait, let's look at the graph again.

Wait, the open circle at \(x=-1\), vertical asymptote at \(x = 1\), jump at \(x = 2\), corner at \(x = 4\), and wait, is there a point at \(x = 2\) and \(x=-1\), \(x = 1\), \(x = 2\), \(x = 4\). Wait, no, let's count again.

Wait, the correct analysis:
- \(x=-1\): Removable discontinuity (not continuous) - not differentiable.
- \(x = 1\): Infinite discontinuity (vertical asymptote) - not differentiable.
- \(x = 2\): Jump discontinuity (not continuous) - not differentiable.
- \(x = 4\): Corner (sharp turn) - not differentiable. Wait, but that's 4 points? Wait, no, maybe I made a mistake. Wait, the options have 5,4,3,2,1. Wait, let's re - check the graph.

Wait, maybe the points are:

1. \(x=-1\) (removable discontinuity)
2. \(x = 1\) (vertical asymptote)
3. \(x = 2\) (jump discontinuity)
4. \(x = 4\) (corner)

Wait, no, maybe I missed one. Wait, the open circle at \(x=-1\), the vertical asymptote at \(x = 1\), the jump at \(x = 2\), the corner at \(x = 4\), and is there a point at \(x = 2\) and another? Wait, no, let's think again.

Wait, the function is not differentiable at:

- \(x=-1\) (removable discontinuity)
- \(x = 1\) (infinite discontinuity)
- \(x = 2\) (jump discontinuity)
- \(x = 4\) (corner)

Wait, that's 4 points? But wait, maybe the correct count is 5? No, let's re - examine the graph description.

Wait, the graph has:

- Open circle at \(x=-1\)
- Vertical asymptote at \(x = 1\)
- Open circle at \(x = 2\) (and closed dot at a different \(y\) - value, jump discontinuity)
- Sharp corner at \(x = 4\)

Wait, maybe I made a mistake. Wait, the answer options include 5,4,3,2,1. Wait, let's recall the standard non - differentiable points:

1. Discontinuities: \(x=-1\) (removable), \(x = 1\) (infinite), \(x = 2\) (jump) - 3 points.
2. Corners: \(x = 4\) - 1 point. Wait, that's 4? Wait, no, maybe there is another point. Wait, maybe the open circle at \(x=-1\), vertical asymptote at \(x = 1\), jump at \(x = 2\), corner at \(x = 4\), and also, maybe the function has a vertical tangent? No, the graph doesn't show a vertical tangent. Wait, maybe the correct number is 5? No, let's re - check.

Wait, let's list all possible non - differentiable points:

- \(x=-1\): Not continuous (removable), not differentiable.
- \(x = 1\): Not continuous (infinite), not differentiable.
- \(x = 2\): Not continuous (jump), not differentiable.
- \(x = 4\): Corner, not differentiable.

Wait, that's 4 points. But let's check the answer options. The option 5 is there, 4 is there. Wait, maybe I missed a point. Wait, maybe at \(x = 2\), there is a jump, at \(x=-1\) removable, \(x = 1\) infinite, \(x = 4\) corner, and another point? Wait, maybe the open circle at \(x=-1\), vertical asymptote at \(x = 1\), jump at \(x = 2\), corner at \(x = 4\), and also, maybe the function has a point where it's not differentiable due to a cusp? No, the graph doesn't show a cusp. Wait, maybe the correct answer is 5? No, let's think again.

Wait, the correct approach:

A function is not differentiable at:

1. Points of discontinuity:
- \(x=-1\): Removable discontinuity.
- \(x = 1\): Infinite discontinuity.
- \(x = 2\): Jump discontinuity.
2. Points of corner:
- \(x = 4\): Corner.

Wait, that's 4 points. But let's check the answer options. The option 4 is present.

Answer:

4