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Step-by-Step Solution
Step 1: Understand the Given Function
We have a piecewise definition for the function
g(x) on the interval [0, 4] :
For 0 \le x \le 3 :
g(x)
= \max_{0 \le t \le x} \{ t^3 - 6t^2 + 9t - 3 \}.
For 3 < x \le 4 :
g(x)
= 4 - x.
We need to find the number of points in the open interval (0,4) where g(x) is not differentiable.
Step 2: Analyze f(t) = t^3 - 6t^2 + 9t - 3
Let us denote
f(t) = t^3 - 6t^2 + 9t - 3.
We first find its derivative:
f'(t) = 3t^2 - 12t + 9 \;=\; 3(t - 1)(t - 3).
The critical points of f are therefore t=1 and t=3 . Next, we check values of f(t) at some key points:
f(0) = -3.
f(1) = 1^3 - 6\cdot1^2 + 9\cdot1 - 3 = 1 - 6 + 9 - 3 = 1.
f(3) = 3^3 - 6\cdot3^2 + 9\cdot3 - 3 = 27 - 54 + 27 - 3 = -3.
From the sign of f'(t) , we see that:
On [0,1] , f'(t) > 0 , so f(t) is increasing from -3 (at t=0 ) to 1 (at t=1 ).
On [1,3] , f'(t) < 0 , so f(t) is decreasing from 1 (at t=1 ) to -3 (at t=3 ).
Step 3: Determine g(x) for 0 \le x \le 3
Because g(x) is defined as \max_{0 \le t \le x}\{f(t)\} , we see:
For 0 \le x \le 1 , the maximum over [0,x] will occur at t = x because f(t) is strictly increasing in [0,1] . Hence,
g(x) = f(x) = x^3 - 6x^2 + 9x - 3.
For 1 \le x \le 3 , since f(t) increases up to t=1 and then decreases, its maximum on [0,x] is f(1) = 1 . Therefore,
g(x) = 1.
Step 4: Determine g(x) for 3 < x \le 4
For 3 < x \le 4 , it is given that
g(x) = 4 - x.
Putting it all together, we have the piecewise definition:
g(x) =
\begin{cases}
x^3 - 6x^2 + 9x - 3, & 0 \le x \le 1, \\
1, & 1 < x \le 3, \\
4 - x, & 3 < x \le 4.
\end{cases}
Step 5: Check Continuity at the Boundaries x=1 and x=3
1. At x=1 :
From the left ( x \to 1^{-} ), g(x) = f(1) = 1 .
From the right ( x \to 1^{+} ), g(x) = 1.
So g is continuous at x=1 .
2. At x=3 :
From the left ( x \to 3^{-} ), g(x) = 1 .
From the right ( x \to 3^{+} ), g(x) = 4 - 3 = 1.
So g is also continuous at x=3 .
Step 6: Check Differentiability Within Each Region
1. For 0 < x < 1 :
g(x) = f(x) , whose derivative is
f'(x) = 3(x - 1)(x - 3).
This is a standard polynomial, differentiable for all x in (0,1) .
2. For 1 < x < 3 :
g(x) = 1 (a constant), whose derivative is 0 , so it is differentiable in that open interval.
3. For 3 < x \le 4 :
g(x) = 4 - x , whose derivative is -1 , so it is differentiable for 3 < x \le 4 .
Step 7: Analyze Differentiability at the “Switch” Points x=1 and x=3
At x=1 :
LHS derivative (from 0 < x < 1 ): f'(1) = 3(1 - 1)(1 - 3) = 0.
RHS derivative (from 1 < x < 3 ): derivative of constant 1 is 0.
Both sides match, so g(x) is differentiable at x=1 .
At x=3 :
LHS derivative (for 1 < x < 3 ): derivative of 1 is 0.
RHS derivative (for 3 < x \le 4 ): derivative of 4 - x is -1.
These do not match ( 0 \neq -1 ), so g(x) is not differentiable at x=3.
Step 8: Conclusion
The only point in the interval (0,4) where g(x) fails to be differentiable is x=3. Hence, the number of such points is
1.
