$f (x) = x^{3} + 5 x^{2} + 6 x + 7$ find $f^{'} (2)$

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Solution

$f (x) = x^{3} + 5 x^{2} + 6 x + 7 f^{'} (x) = 3 x^{2} + 10 x + 6 f^{'} (2) = 3 (4) + 10 (2) + 6 12 + 20 + 6 = 38$

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f^{'} (3)

f (x) = x^{3} + 5 x^{2} - 3 x + 5

f (x) = x^{3} + 7 x^{2} + 8 x - 9

f (x) = x^{3} - {5 x}^{2} - 3 x

\sqrt{5 x^{2} - 6 x + 8} - \sqrt{5 x^{2} - 6 x - 7} = 1

f (x) = \frac{x^{4}}{4} + \frac{2}{3} x^{3} - \frac{5}{2} x^{2} - 6 x + 7

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