Answer: choice C
Explanation
Use Desmos or a similar tool to plot the graph of y = x^2 - 9.
It will make a parabola with the x intercepts at -3 and 3.
Between those roots, the parabola is below the x axis. When applying absolute value, that portion will mirror over to land above the x axis. Graph y = |x^2-9| to see this mirroring take place. The resulting curve should look like the letter W.
The graph of y = |x^2-9| has sharp cusp points when x = -3 and x = 3. These x values lead to "not differentiable". Why are these cusp points? Because they are stationary points that didn't move during the reflection, and they are on the boundary of the reflected portion.
In short, reflecting breaks the differentiability at x = -3 and x = 3 since that small little portion is no longer a smooth transition.
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Another approach would be to use tangent slopes.
As you get closer and closer to x = -3 from the left, the tangent slopes will be negative. Once you pass by x = -3, the tangent slopes will suddenly become positive. This jump discontinuity in f ' (x) will mean f(x) is not differentiable here.
Similar logic applies for x = 3 as well.
That situation does not happen for x = 0 or x = 9.
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If you are familiar and comfortable with using derivatives, then we can say the following
[tex]f( \text{x} ) = |\text{x}^2 - 9| \\\\f'( \text{x} ) = \frac{2 \text{x}|\text{x}^2 - 9|}{\text{x}^2 - 9} \\\\[/tex]
Notice how x = -3 or x = 3 will make the denominator of f ' (x) to be zero, which makes f ' (x) not differentiable here.
x = 0 does not lead to a division by zero error. Neither does x = 9.
