1. Use conservation of mechanical energy

For one-dimensional motion under a conservative force, $E_{\text{mech}} = K + U(x).$ Given $E_{\text{mech}} = 8\,\text{J},$ so the kinetic energy at position $x$ is $K(x) = 8 - U(x).$

Thus:

• where $U$ is minimum, $K$ is maximum,
• where $U$ is maximum, $K$ is minimum,
• if $U$ is constant in some region, then $K$ is also constant there.

2. Read the graph qualitatively

From the given potential-energy graph:

• for $x > x_4$, the graph is horizontal, so $U(x)$ is constant there,
• at $x=x_2$, $U$ is minimum,
• at $x=x_3$, the graph shows $U(x_3)=4\,\text{J}$,
• for $x < x_1$, the potential is not the maximum value of the graph; hence kinetic energy there is not the smallest.

3. Check each option

Option A

For $x>x_4$, $U(x)$ is constant. Therefore, $K = 8 - U = \text{constant}.$ So the kinetic energy is constant throughout that region.

✅ A is correct.

Option B

Statement: at $x<x_1$, K.E. is smallest and the particle is moving at the slowest speed.

Kinetic energy is smallest where potential energy is largest, because $K=8-U.$ From the graph, the largest relevant potential is not in the region $x<x_1$. Hence the statement that KE is smallest for $x<x_1$ is false.

❌ B is incorrect.

Option C

At $x=x_2$, the potential energy is minimum. Hence $K=8-U(x_2)$ is maximum there. Maximum kinetic energy means maximum speed.

✅ C is correct.

Option D

At $x=x_3$, from the graph, $U(x_3)=4\,\text{J}.$ Therefore, $K(x_3)=8-4=4\,\text{J}.$

✅ D is correct.

4. Conclusion

The incorrect statement is: $\boxed{\text{B}}$