Answer :
The value of 'a' that makes F(x) an antiderivative of f(x) = 42x⁵ is a = 7.
To determine the value of 'a' that makes F(x) an antiderivative of f(x), we need to find the derivative of F(x) and check if it matches f(x).
Given that f(x) = 42x⁵ and F(x) = ax⁶ + 35x⁴ + 42x⁷, we can find the derivative of F(x) by differentiating each term separately.
The derivative of ax⁶ is 6ax⁵ (using the power rule).
The derivative of 35x⁴ is 140x³ (using the power rule).
The derivative of 42x⁷ is 294x⁶ (using the power rule).
Now, let's compare the derivative of F(x) with f(x) = 42x⁵.
6ax⁵+ 140x³ + 294x⁶ = 42x⁵
To match the terms on both sides of the equation, we can equate the corresponding coefficients.
6a = 42 (for the x⁵ term)
140 = 0 (for the x³ term)
294 = 0 (for the x⁶ term)
Simplifying the equation, we find:
6a = 42
Dividing both sides by 6:
a = 7
Therefore, the value of 'a' that makes F(x) an antiderivative of f(x) = 42x⁵ is a = 7.
In summary, to make F(x) an antiderivative of f(x) = 42x⁵, the value of 'a' needs to be 7.
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