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Step-by-Step Solution
Step 1: Express the function in a suitable piecewise form
The given function is
f(x) = |(x - 1)((x)^2 - 2x - 3)| + (x - 3) .
Factorize (x)^2 - 2x - 3 as (x + 1)(x - 3) . Hence,
(x - 1)((x)^2 - 2x - 3) = (x - 1)(x + 1)(x - 3) .
We need to consider the sign of this product over the interval (0,4) to determine how the absolute value behaves.
Step 2: Determine intervals based on sign changes
The expression (x - 1)(x + 1)(x - 3) has zeros at x = 1,\,-1,\,3 . Within (0,4) , the relevant sign changes occur at x=1 and x=3 . We split the interval (0,4) into three subintervals:
0 < x < 1
1 \le x < 3
3 \le x \le 4
In each interval, determine whether (x - 1)(x + 1)(x - 3) is positive or negative to remove the absolute value accordingly.
Step 3: Rewrite f(x) in each subinterval
• For 0 < x < 1 : The term (x - 1) is negative, (x + 1) is positive, (x - 3) is negative. Overall, the product (x - 1)(x + 1)(x - 3) is positive × negative × negative, which is positive. Thus,
|(x - 1)(x + 1)(x - 3)| = (x - 1)(x + 1)(x - 3) .
However, observe carefully that (x - 1) is negative and (x - 3) is negative, so their product is positive. Indeed, the entire product is positive in the interval (0,1) . Hence
f(x) = (x - 1)(x + 1)(x - 3) + (x - 3) ,
which simplifies to
f(x) = (x - 3)x^2
(because (x - 1)(x + 1) = x^2 - 1 , but we will rely on the provided piecewise form for consistency).
• For 1 \le x < 3 : The factor (x - 1) is nonnegative, (x + 1) is positive, (x - 3) is negative. Hence the overall product is negative or zero because only one factor (namely (x - 3) ) is negative for 1 < x < 3 . Therefore,
|(x - 1)(x + 1)(x - 3)| = -\,[(x - 1)(x + 1)(x - 3)].
This leads to
f(x) = - \, (x - 1)(x + 1)(x - 3) + (x - 3) .
With the provided piecewise form, this is given as
f(x) = (x - 3)\bigl(2 - x^2\bigr).
• For 3 \le x \le 4 : Now (x - 1) and (x - 3) are nonnegative, and (x + 1) is positive, making the product (x - 1)(x + 1)(x - 3) nonnegative. Thus, within [3,4] ,
|(x - 1)(x + 1)(x - 3)| = (x - 1)(x + 1)(x - 3).
Hence
f(x) = (x - 3)x^2,
consistent with the piecewise expression.
Step 4: Compute the derivative in each interval
Based on the piecewise form, the derivative f'(x) in each subinterval is:
For 0 < x < 1 : f(x) = (x - 3)x^2 \implies f'(x) = 3x^2 - 6x.
For 1 \le x < 3 : f(x) = (x - 3)\bigl(2 - x^2\bigr) \implies f'(x) = -3x^2 + 6x + 2.
For 3 \le x \le 4 : f(x) = (x - 3)x^2 \implies f'(x) = 3x^2 - 6x.
Step 5: Check critical points and sign changes at boundaries x=1 and x=3
- At x = 1 :
• Just to the left ( x < 1 , in (0,1) ), f'(x) = 3x^2 - 6x . Substituting a value slightly less than 1 (say 0.9 ) would give a negative derivative (because 3(0.9)^2 - 6(0.9) < 0 ).
• Just to the right ( 1 < x < 3 ), f'(x) = -3x^2 + 6x + 2 . At x = 1^+ , this is -3(1)^2 + 6(1) + 2= -3 + 6 + 2 = 5 > 0.
The derivative changes from negative to positive, indicating a local minimum at x = 1 .
- At x = 3 :
• Just to the left ( 1 < x < 3 ), f'(x) = -3x^2 + 6x + 2 . At x = 3^- , f'(3^-) = -3(9) + 18 + 2= -27 + 20= -7 < 0.
• Just to the right ( 3 < x < 4 ), f'(x) = 3x^2 - 6x . At x = 3^+ , f'(3^+) = 3(9) - 18= 27 - 18= 9 > 0.
The derivative changes from negative to positive, indicating a local minimum at x = 3 .
Step 6: Check for any critical points inside the intervals
- In (1,3) : The derivative is f'(x) = -3x^2 + 6x + 2 . Setting f'(x) = 0 gives a quadratic equation:
-3x^2 + 6x + 2 = 0.
Solve using the quadratic formula:
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} with a = -3 , b = 6 , c = 2 .
So:
x = \frac{-6 \pm \sqrt{6^2 - 4(-3)(2)}}{2(-3)}
= \frac{-6 \pm \sqrt{36 + 24}}{-6}
= \frac{-6 \pm \sqrt{60}}{-6}
= \frac{-6 \pm \sqrt{4 \cdot 15}}{-6}
= \frac{-6 \pm 2\sqrt{15}}{-6}.
At least one real solution lies in (1,3) . Checking the sign change of the derivative around this root shows that this root corresponds to a local maximum.
- In (0,1) : f'(x) = 3x^2 - 6x . Setting this to zero gives 3x^2 - 6x = 0 \implies 3x(x - 2)= 0, which has solutions x=0 and x=2 . Neither falls strictly inside (0,1) , so there is no internal critical point in (0,1) that causes a local extremum.
- In (3,4) : f'(x) = 3x^2 - 6x . The solutions to 3x^2 - 6x =0 are again x=0 or x=2 , so no new critical points in (3,4) .
Step 7: Count the local minima and maxima
From the analysis:
Local minima at x=1 and x=3 .
One local maximum in the interval (1,3) due to the derivative equating to zero there.
No other local extrema in (0,1) or (3,4) .
Therefore, the number of points of local minimum is m = 2 and the number of points of local maximum is M = 1 .
Step 8: Final answer
The sum m + M = 2 + 1 = 3.
