1. Given function

We need the number of points where $f(x)=|x|+|x^2-1|$ attains either a local maximum or a local minimum.

So first, we remove the modulus signs by splitting the real line into intervals.

2. Break into cases

The critical sign-change points are:

• $x=0$ from $|x|$
• $x=\pm 1$ from $|x^2-1|$

Now analyze interval-wise.

Case 1: $x\le -1$

Here,

• $|x|=-x$
• $x^2-1\ge 0 \Rightarrow |x^2-1|=x^2-1$

So, $f(x)=-x+x^2-1=x^2-x-1$

Then, $f'(x)=2x-1$ For $x\le -1$, we have $2x-1<0$, so $f$ is decreasing on $(-\infty,-1]$.

Case 2: $-1\le x\le 0$

Here,

• $|x|=-x$
• $x^2-1\le 0 \Rightarrow |x^2-1|=1-x^2$

So, $f(x)=-x+1-x^2=1-x-x^2$

Then, $f'(x)=-1-2x$ Set $f'(x)=0$: $-1-2x=0 \Rightarrow x=-\frac12$

Also,

• for $x< -\frac12$, $f'(x)>0$
• for $x> -\frac12$, $f'(x)<0$

Hence at $x=-\frac12$ $f$ changes from increasing to decreasing, so this is a local maximum.

Case 3: $0\le x\le 1$

Here,

• $|x|=x$
• $|x^2-1|=1-x^2$

So, $f(x)=x+1-x^2=1+x-x^2$

Then, $f'(x)=1-2x$ Set $f'(x)=0$: $1-2x=0 \Rightarrow x=\frac12$

Also,

• for $x<\frac12$, $f'(x)>0$
• for $x>\frac12$, $f'(x)<0$

Hence at $x=\frac12$ $f$ changes from increasing to decreasing, so this is a local maximum.

Case 4: $x\ge 1$

Here,

• $|x|=x$
• $|x^2-1|=x^2-1$

So, $f(x)=x+x^2-1=x^2+x-1$

Then, $f'(x)=2x+1$ For $x\ge 1$, $2x+1>0$, so $f$ is increasing on $[1,\infty)$.

3. Check the junction points

Now examine $x=-1,0,1$ carefully.

At $x=-1$

• On $(-\infty,-1)$, $f$ is decreasing.
• On $(-1,0)$, near $-1$, $f'(x)=-1-2x>0$ (for example at $x=-0.9$, this is $0.8>0$), so $f$ is increasing just to the right of $-1$.

Thus, decreasing to the left and increasing to the right means $x=-1$ is a local minimum.

At $x=0$

• On $(-1,0)$, near $0$, $f'(x)=-1-2x<0$, so decreasing from the left.
• On $(0,1)$, near $0$, $f'(x)=1-2x>0$, so increasing to the right.

Thus, decreasing to the left and increasing to the right means $x=0$ is a local minimum.

At $x=1$

• On $(0,1)$, near $1$, $f'(x)=1-2x<0$, so decreasing from the left.
• On $(1,\infty)$, $f'(x)=2x+1>0$, so increasing to the right.

Thus, decreasing to the left and increasing to the right means $x=1$ is a local minimum.

4. List all local extrema points

• Local minima at: $x=-1,\;0,\;1$
• Local maxima at: $x=-\frac12,\;\frac12$

So total number of points is $3+2=5$

5. Comparison with stored answer

Derived answer = $5$.

Stored correct answer = $5$.

Hence, the derived answer agrees with the stored answer.