To solve the problem, we need to follow the steps outlined in parts (a), (b), and (c).
Part (a)
First, we need to find \( f(a + h) - f(a)h \) for the function \( f(x) = 2x^2 - 7x + 25 \).
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Calculate \( f(a + h) \): \[ f(a + h) = 2(a + h)^2 - 7(a + h) + 25 \] Expanding \( (a + h)^2 \): \[ (a + h)^2 = a^2 + 2ah + h^2 \] Therefore, \[ f(a + h) = 2(a^2 + 2ah + h^2) - 7(a + h) + 25 = 2a^2 + 4ah + 2h^2 - 7a - 7h + 25 \] This simplifies to: \[ f(a + h) = 2a^2 - 7a + 25 + 4ah - 7h + 2h^2 \] Here, \( f(a) = 2a^2 - 7a + 25 \).
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Calculate \( f(a + h) - f(a)h \): We first compute \( f(a)h \): \[ f(a)h = (2a^2 - 7a + 25)h \] Thus, \[ f(a + h) - f(a)h = \left( 2a^2 - 7a + 25 + 4ah - 7h + 2h^2 \right) - (2a^2 - 7a + 25)h \] Simplifying this: \[ f(a + h) - f(a)h = (2a^2 - 7a + 25) + 4ah - 7h + 2h^2 - (2a^2 - 7a + 25)h \] \[ = 2a^2 - 7a + 25 - (2a^2 - 7a + 25)h + 4ah - 7h + 2h^2 \] \[ = (2a^2 - 7a + 25)(1 - h) + 4ah - 7h + 2h^2 \]
Therefore, the final expression may be written as: \[ f(a + h) - f(a)h = (2a^2 - 7a + 25)(1 - h) + 4ah - 7h + 2h^2 \]
Part (b)
Now, computing the limit: \[ \lim_{h \to 0} f(a + h) - f(a)h \]
As \( h \) approaches 0, \( (1 - h) \) approaches 1: \[ \lim_{h \to 0} (2a^2 - 7a + 25)(1 - h) + 4ah - 7h + 2h^2 \] The terms involving \( h \) vanish: \[ = 2a^2 - 7a + 25 \]
Part (c)
To find \( f'(4) \): \[ f'(4) = \lim_{h \to 0} \frac{f(4 + h) - f(4)}{h} \] Using the result from part (b), plug in \( a = 4 \): \[ f'(4) = 2(4^2) - 7(4) + 25 = 2(16) - 28 + 25 = 32 - 28 + 25 = 29 \]
Thus, the answers for parts (a), (b), and (c) are:
- (a) \( f(a + h) - f(a)h \)
- (b) \( 2a^2 - 7a + 25 \)
- (c) \( f'(4) = 29 \)
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