Find the x-coordinates of all relative maxima of f(x) f(x)=-x^3-9x^2+81x-16 Answer Attemptiout of Additional Solution (C) No Solution x=

To find the x-coordinates of the relative maxima of the function $f(x)=-x^{3}-9x^{2}+81x-16$, we need to find the critical points of the function and then determine which of these points correspond to relative maxima.<br /><br />Step 1: Find the derivative of the function $f(x)$.<br />The derivative of $f(x)$ is given by:<br />$f'(x) = -3x^2 - 18x + 81$<br /><br />Step 2: Set the derivative equal to zero and solve for $x$.<br />$-3x^2 - 18x + 81 = 0$<br />$-3(x^2 + 6x - 27) = 0$<br />$-3(x + 9)(x - 3) = 0$<br />$x = -9$ or $x = 3$<br /><br />Step 3: Determine the second derivative of the function $f(x)$.<br />The second derivative of $f(x)$ is given by:<br />$f''(x) = -6x - 18$<br /><br />Step 4: Evaluate the second derivative at the critical points.<br />$f''(-9) = -6(-9) - 18 = 54 - 18 = 36 > 0$<br />$f''(3) = -6(3) - 18 = -18 - 18 = -36 < 0$<br /><br />Since $f''(-9) > 0$, the critical point $x = -9$ corresponds to a relative minimum.<br />Since $f''(3) < 0$, the critical point $x = 3$ corresponds to a relative maximum.<br /><br />Therefore, the x-coordinate of the relative maximum of the function $f(x)=-x^{3}-9x^{2}+81x-16$ is $x = 3$.