Find the x -coordinates of all relative maxima of f(x) - f(x)=3x^3+45x^2+81x+11 Answer Attempt 1 out of 4 x=

To find the x-coordinates of all relative maxima of the function $f(x) = 3x^3 + 45x^2 + 81x + 11$, 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 />$f'(x) = 9x^2 + 90x + 81$<br /><br />Step 2: Set the derivative equal to zero and solve for $x$.<br />$9x^2 + 90x + 81 = 0$<br />$(x + 9)(x + 9) = 0$<br />$x = -9$<br /><br />Step 3: Determine the second derivative of the function $f(x)$.<br />The second derivative of $f(x)$ is given by:<br />$f''(x) = 18x + 90$<br /><br />Step 4: Evaluate the second derivative at the critical point $x = -9$.<br />$f''(-9) = 18(-9) + 90 = -162 + 90 = -72$<br /><br />Since $f''(-9) < 0$, the critical point $x = -9$ corresponds to a relative maximum.<br /><br />Therefore, the x-coordinate of the relative maximum of the function $f(x) = 3x^3 + 45x^2 + 81x + 11$ is $x = -9$.