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Find f '(−3), if f(x) = (2x2 − 7x)(−x2 − 7). Round your answer to the nearest integer. Use the hyphen symbol, -, for negative values.
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\(\large\color{black}{ \displaystyle f(x)=(2x^2-7x)(-x^2-7) }\)
\(\large\color{black}{ \displaystyle f'(x)=(2x^2-7x)'\times (-x^2-7)+(2x^2-7x)\times (-x^2-7)' }\)
\(\large\color{black}{ \displaystyle (2x^2-7x)' = \quad ?\\[0.7em] (-x^2-7)' = \quad ? }\)
By \('\) I am denoting [the first} derivative.
Then, when you find \(\large\color{black}{ \displaystyle f'(x) }\), plug in x=3.
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You can find the f`(x) either using the product rule I proposed, or you can first expand and then differentiate. That doesn't matter.
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