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To find the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] so that the function [tex]\( f(x) \)[/tex] satisfies the hypotheses of the Mean Value Theorem (MVT) on the interval [tex]\([0,2]\)[/tex], we must ensure that [tex]\( f(x) \)[/tex] is continuous and differentiable on the interval [tex]\([0,2]\)[/tex]. Here is a step-by-step solution:
1. Continuity at [tex]\( x = 1 \)[/tex]:
For [tex]\( f(x) \)[/tex] to be continuous at [tex]\( x = 1 \)[/tex], the values from the two pieces of the function must match at [tex]\( x = 1 \)[/tex]. Thus,
[tex]\[ 7(1)^2 - 3(1) + a = b(1) + c \][/tex]
Simplifying,
[tex]\[ 7 - 3 + a = b + c \implies 4 + a = b + c \quad \text{(Equation 1)} \][/tex]
2. Continuity at [tex]\( x = 0 \)[/tex]:
For [tex]\( f(x) \)[/tex] to be continuous at [tex]\( x = 0 \)[/tex],
[tex]\[ f(0) = 9 \][/tex]
Since [tex]\( x = 0 \)[/tex] already has [tex]\( f(0) = 9 \)[/tex], no further conditions are needed here because [tex]\( f(x) \)[/tex] is already defined correctly.
3. Continuity at [tex]\( x = 2 \)[/tex]:
For [tex]\( f(x) \)[/tex] to be continuous at [tex]\( x = 2 \)[/tex],
[tex]\[ b(2) + c = f(2) \][/tex]
Since [tex]\( x = 2 \)[/tex] falls within the linear part of the piecewise function (where [tex]\( 1 \leq x \leq 2 \)[/tex]), we don't have another piece to match to but need to determine [tex]\( c \)[/tex] in relation to [tex]\( b \)[/tex] for simplicity.
4. Differentiability at [tex]\( x = 1 \)[/tex]:
For [tex]\( f(x) \)[/tex] to be differentiable at [tex]\( x = 1 \)[/tex], the derivatives from both sides of [tex]\( x = 1 \)[/tex] must be equal. First, we need to calculate the derivative of each piece:
- For [tex]\( 0 < x < 1 \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}(7x^2 - 3x + a) = 14x - 3 \][/tex]
- For [tex]\( 1 \leq x \leq 2 \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}(bx + c) = b \][/tex]
At [tex]\( x = 1 \)[/tex], derivatives must match:
[tex]\[ f'_-(1) = f'_+(1) \][/tex]
[tex]\[ 14(1) - 3 = b \implies 14 - 3 = b \implies b = 11 \quad \text{(Equation 2)} \][/tex]
5. Substitute [tex]\( b \)[/tex] back into Equation 1 from Step 1:
[tex]\[ 4 + a = 11 + c \][/tex]
Simplifying further,
[tex]\[ a - c = 7 \quad \text{(Equation 3)} \][/tex]
6. Express [tex]\( c \)[/tex] in terms of [tex]\( a \)[/tex] (since we now have Equation 3):
[tex]\[ c = a - 7 \][/tex]
Putting it all together, we have:
- Equation 2: [tex]\( b = 11 \)[/tex]
- Equation 3 and continuity adjustment: [tex]\( c = a - 7 \)[/tex]
Finally, choosing a value for [tex]\( a \)[/tex] that maintains the continuity and differentiability:
- Let [tex]\( a = \boxed{16} \)[/tex]:
Then,
- [tex]\( b = \boxed{11} \)[/tex]
- [tex]\( c = a - 7 = 16 - 7 = \boxed{9} \)[/tex].
Hence, the values that satisfy the MVT conditions are [tex]\( a = 16 \)[/tex], [tex]\( b = 11 \)[/tex], and [tex]\( c = 9 \)[/tex].
1. Continuity at [tex]\( x = 1 \)[/tex]:
For [tex]\( f(x) \)[/tex] to be continuous at [tex]\( x = 1 \)[/tex], the values from the two pieces of the function must match at [tex]\( x = 1 \)[/tex]. Thus,
[tex]\[ 7(1)^2 - 3(1) + a = b(1) + c \][/tex]
Simplifying,
[tex]\[ 7 - 3 + a = b + c \implies 4 + a = b + c \quad \text{(Equation 1)} \][/tex]
2. Continuity at [tex]\( x = 0 \)[/tex]:
For [tex]\( f(x) \)[/tex] to be continuous at [tex]\( x = 0 \)[/tex],
[tex]\[ f(0) = 9 \][/tex]
Since [tex]\( x = 0 \)[/tex] already has [tex]\( f(0) = 9 \)[/tex], no further conditions are needed here because [tex]\( f(x) \)[/tex] is already defined correctly.
3. Continuity at [tex]\( x = 2 \)[/tex]:
For [tex]\( f(x) \)[/tex] to be continuous at [tex]\( x = 2 \)[/tex],
[tex]\[ b(2) + c = f(2) \][/tex]
Since [tex]\( x = 2 \)[/tex] falls within the linear part of the piecewise function (where [tex]\( 1 \leq x \leq 2 \)[/tex]), we don't have another piece to match to but need to determine [tex]\( c \)[/tex] in relation to [tex]\( b \)[/tex] for simplicity.
4. Differentiability at [tex]\( x = 1 \)[/tex]:
For [tex]\( f(x) \)[/tex] to be differentiable at [tex]\( x = 1 \)[/tex], the derivatives from both sides of [tex]\( x = 1 \)[/tex] must be equal. First, we need to calculate the derivative of each piece:
- For [tex]\( 0 < x < 1 \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}(7x^2 - 3x + a) = 14x - 3 \][/tex]
- For [tex]\( 1 \leq x \leq 2 \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}(bx + c) = b \][/tex]
At [tex]\( x = 1 \)[/tex], derivatives must match:
[tex]\[ f'_-(1) = f'_+(1) \][/tex]
[tex]\[ 14(1) - 3 = b \implies 14 - 3 = b \implies b = 11 \quad \text{(Equation 2)} \][/tex]
5. Substitute [tex]\( b \)[/tex] back into Equation 1 from Step 1:
[tex]\[ 4 + a = 11 + c \][/tex]
Simplifying further,
[tex]\[ a - c = 7 \quad \text{(Equation 3)} \][/tex]
6. Express [tex]\( c \)[/tex] in terms of [tex]\( a \)[/tex] (since we now have Equation 3):
[tex]\[ c = a - 7 \][/tex]
Putting it all together, we have:
- Equation 2: [tex]\( b = 11 \)[/tex]
- Equation 3 and continuity adjustment: [tex]\( c = a - 7 \)[/tex]
Finally, choosing a value for [tex]\( a \)[/tex] that maintains the continuity and differentiability:
- Let [tex]\( a = \boxed{16} \)[/tex]:
Then,
- [tex]\( b = \boxed{11} \)[/tex]
- [tex]\( c = a - 7 = 16 - 7 = \boxed{9} \)[/tex].
Hence, the values that satisfy the MVT conditions are [tex]\( a = 16 \)[/tex], [tex]\( b = 11 \)[/tex], and [tex]\( c = 9 \)[/tex].
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