F(x) = =
Find f'(x) and f''(x).
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ALEHER
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f''(x) = {
Epe

To make things easier, first rewrite

f(x) = x / (x + 7/x) = x ² / (x ² + 7)

i.e. multiply f by x/x, which is valid because x can't be 0 anyway as it's outside the domain of f.

Now, by the quotient and chain rules, we get

f '(x) = ((x ² + 7) (x ²)' - x ² (x ² + 7)') / (x ² + 7)²

… = (2x (x ² + 7) - x ² (2x)) / (x ² + 7)²

… = (2x ³ + 14x - 2x ³) / (x ² + 7)²

… = 14x / (x ² + 7)²

f ''(x) = ((x ² + 7)² (14x)' - (14x) ((x ² + 7)²)') / ((x ² + 7)²)²

… = (14 (x ² + 7)² - (14x) (2 (x ² + 7)² (x ² + 7)')) / (x ² + 7)⁴

… = (14 (x ² + 7)² - (14x) (2 (2x) (x ² + 7)²)) / (x ² + 7)⁴

… = (14 (x ² + 7)² - 4x (14x) (x ² + 7)²) / (x ² + 7)⁴

… = 14 (x ² + 7)² (1 - 4x) / (x ² + 7)⁴

… = 14 (1 - 4x) / (x ² + 7)²

… = (14 - 56x) / (x ² + 7)²

answer: domain and range mean all possible values of x and y that could be substituted to the inequality. a solution means all possible values that make the inequality statement true. so while the range is still all real numbers, the solution is the shaded area, for which y will always be greater than the line of the function.

answer: sine

step-by-step explanation: