Find the x -coordinates of all relative maxima of f(x) f(x)=-(3)/(4)x^4-3x^3+81x^2-37 Answer Attempt 3 out of 4 (4) Additional Solution (- No Solution x=

To find the x-coordinates of all relative maxima of the function $f(x)=-\frac{3}{4}x^{4}-3x^{3}+81x^{2}-37$, 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 />$f'(x) = -3x^{3} - 9x^{2} + 162x$<br /><br />Step 2: Set the derivative equal to zero and solve for $x$.<br />$-3x^{3} - 9x^{2} + 162x = 0$<br />$-3x(x^{2} + 3x - 54) = 0$<br />$-3x(x + 9)(x - 6) = 0$<br /><br />The solutions to this equation are $x = 0$, $x = -9$, and $x = 6$.<br /><br />Step 3: Determine which of these critical points correspond to relative maxima.<br />To determine which of these critical points correspond to relative maxima, we need to evaluate the second derivative of the function at each critical point.<br /><br />The second derivative of $f(x)$ is given by:<br />$f''(x) = -9x^{2} - 18x + 162$<br /><br />Now, let's evaluate the second derivative at each critical point:<br /><br />For $x = 0$:<br />$f''(0) = -9(0)^{2} - 18(0) + 162 = 162$<br /><br />For $x = -9$:<br />$f''(-9) = -9(-9)^{2} - 18(-9) + 162 = -729 + 162 + 162 = -405$<br /><br />For $x = 6$:<br />$f''(6) = -9(6)^{2} - 18(6) + 162 = -324 - 108 + 162 = -270$<br /><br />Since $f''(0) > 0$, the critical point $x = 0$ corresponds to a relative minimum.<br /><br />Since $f''(-9) < 0$, the critical point $x = -9$ corresponds to a relative maximum.<br /><br />Since $f''(6) < 0$, the critical point $x = 6$ corresponds to a relative maximum.<br /><br />Therefore, the x-coordinates of all relative maxima of the function $f(x)=-\frac{3}{4}x^{4}-3x^{3}+81x^{2}-37$ are $x = -9$ and $x = 6$.