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Step-by-Step Solution
Step 1: Assume the general form of the cubic polynomial
Let
f(x) = ax^3 + bx^2 + cx + d
where a, b, c, and d are real constants.
Step 2: Apply the given values for f(x)
1) Given
f(-1) = 10 .
Substitute
x = -1
into
f(x) :
a(-1)^3 + b(-1)^2 + c(-1) + d = -a + b - c + d = 10 \quad (1)
2) Given
f(1) = -6 .
Substitute
x = 1
into
f(x) :
a(1)^3 + b(1)^2 + c(1) + d = a + b + c + d = -6 \quad (2)
Add equations (1) and (2):
(-a + b - c + d) + (a + b + c + d) = 10 + (-6)
2b + 2d = 4 \quad \Rightarrow \quad b + d = 2 \quad (3)
Step 3: Use information about critical points of f(x)
The first derivative of
f(x)
is
f'(x) = 3ax^2 + 2bx + c .
We are told
x = -1
is a critical point for
f(x) ,
so
f'(-1) = 0 :
3a(-1)^2 + 2b(-1) + c = 3a - 2b + c = 0 \quad (4)
Step 4: Use information about the critical point of f'(x)
The second derivative of
f(x)
is
f''(x) = 6ax + 2b .
Given that
x = 1
is a critical point of
f'(x)
implies
f''(1) = 0 :
6a(1) + 2b = 0 \quad \Rightarrow \quad 6a + 2b = 0 \quad \Rightarrow \quad b = -3a \quad (5)
Step 5: Relate constants using the derivatives
From equation (4) and substituting
b = -3a
into
3a - 2b + c = 0 :
3a - 2(-3a) + c = 3a + 6a + c = 9a + c = 0 \quad \Rightarrow \quad c = -9a.
Step 6: Express f(x) in terms of a single constant
We have:
a = a
b = -3a
c = -9a
From
b + d = 2
and
b = -3a ,
we have
d = 2 - b = 2 + 3a.
So,
f(x) = a x^3 - 3a x^2 - 9a x + (2 + 3a).
Step 7: Use the condition f(1) = -6 again
Alternatively, we can also simplify directly with
a + b + c + d = -6
and substitute
a = a, \; b = -3a, \; c = -9a, \; d = 2 + 3a :
a + (-3a) + (-9a) + (2 + 3a) = -6
a - 3a - 9a + 2 + 3a = -6
(a - 3a - 9a + 3a) + 2 = -6
-8a + 2 = -6 \quad \Rightarrow \quad -8a = -8 \quad \Rightarrow \quad a = 1.
Then
b = -3a = -3 \times 1 = -3
c = -9a = -9
d = 2 + 3a = 2 + 3 = 5
Thus,
f(x) = x^3 - 3x^2 - 9x + 5.
Step 8: Find the critical points of f(x)
Compute
f'(x) :
f'(x) = 3x^2 - 6x - 9.
For possible local maxima or minima, set
f'(x) = 0 :
3x^2 - 6x - 9 = 0
Divide through by 3:
x^2 - 2x - 3 = 0
This factors as:
(x - 3)(x + 1) = 0 \quad \Rightarrow \quad x = 3 \; \text{or} \; x = -1.
Step 9: Determine which critical point is a minima
Compute
f''(x) :
f''(x) = 6x - 6.
Evaluate at
x = 3 :
f''(3) = 6(3) - 6 = 18 - 6 = 12 > 0,
which indicates a local minimum at
x = 3.
Evaluate at
x = -1 :
f''(-1) = 6(-1) - 6 = -6 - 6 = -12 < 0,
which indicates a local maximum at
x = -1.
Final Answer
The polynomial
f(x)
has a local minimum at
x = 3.
