Question

The point of inflection of the function $f(x)=x^3$ in the interval $[-1,1]$ is

Hint:

To find the point of inflection, first compute the second derivative and then check whether its sign changes at the critical value.

Answer 1

Step 1: Recall the definition of a point of inflection.
A point of inflection is a point on the graph where the concavity of the function changes. This occurs when the second derivative of the function changes its sign. 

Step 2: Compute the first derivative. 
The given function is \[ f(x)=x^3 \] Differentiate once \[ f'(x)=3x^2 \] 
Step 3: Compute the second derivative. 
Differentiate again \[ f''(x)=6x \] 
Step 4: Determine where the second derivative is zero. 
\[ 6x=0 \] \[ x=0 \] 
Step 5: Check change of concavity. 
For $x<0$ \[ f''(x)<0 \] For $x>0$ \[ f''(x)>0 \] Thus the concavity changes from downward to upward at $x=0$. 

Step 6: Conclusion. 
Therefore the point of inflection occurs at \[ x=0 \] which lies within the interval $[-1,1]$. 
Final Answer: $\boxed{0}$ 

"