Question

If \(f(x) = \min\{1, x^2, x^3\}\), then

• A. \(f(x)\) is not everywhere continuous
• B. \(f(x)\) is continuous and differentiable everywhere
• C. \(f(x)\) is not differentiable at two points
• D. \(f(x)\) is not differentiable at one point

Hint:

For \(\min\) functions, non-differentiability occurs at points where the minimum function changes.

Answer 1

The Correct Option is D

Step 1:
Find the points where the functions intersect:

\[ x^2 = x^3 \Rightarrow x^2(x-1)=0 \Rightarrow x=0,1 \] \[ x^2 = 1 \Rightarrow x=\pm1 \] \[ x^3 = 1 \Rightarrow x=1 \] 
So, the critical points are: \[ x = -1, 0, 1 \]

Step 2:
Determine the function in different intervals:

For \( x < 0 \):
\[ f(x) = x^3 \] For \( 0 \le x \le 1 \):
\[ f(x) = x^3 \] For \( x > 1 \):
\[ f(x) = 1 \] 
Hence, the function can be written as: 

\[ f(x) = \begin{cases} x^3, & x \le 1 \\ 1, & x \ge 1 \end{cases} \]

Step 3:
Check differentiability at \( x = 1 \):

Left-hand derivative (LHD): \[ \frac{d}{dx}(x^3) = 3x^2 \Rightarrow 3(1)^2 = 3 \] Right-hand derivative (RHD): \[ \frac{d}{dx}(1) = 0 \] 
Since, \[ \text{LHD} \ne \text{RHD} \] 
Therefore, the function is not differentiable at \( x = 1 \).

Final Conclusion:
\[ \boxed{\text{The function is not differentiable at one point (}x=1\text{)}} \]