Find the x-coordinates of all relative maxima of f(x) f(x)=(3)/(5)x^5+12x^4+60x^3+3 Answer Attempticut of 4 (C) Additional Solution No Solution z=1

To find the x-coordinates of all relative maxima of the function $f(x) = \frac{3}{5}x^5 + 12x^4 + 60x^3 + 3$, we need to find the critical points of the function and then determine which of those points correspond to relative maxima.<br /><br />Step 1: Find the derivative of $f(x)$.<br />The derivative of $f(x)$ is given by:<br />$f'(x) = \frac{d}{dx} \left( \frac{3}{5}x^5 + 12x^4 + 60x^3 + 3 \right)$<br />$f'(x) = 3x^4 + 48x^3 + 180x^2$<br /><br />Step 2: Set the derivative equal to zero and solve for $x$.<br />$3x^4 + 48x^3 + 180x^2 = 0$<br />$3x^2(x^2 + 16x + 60) = 0$<br />$x^2(x + 10)(x + 6) = 0$<br /><br />The solutions to this equation are:<br />$x = 0, x = -10, x = -6$<br /><br />Step 3: Determine which of these points correspond to relative maxima.<br />To determine which of these points correspond to relative maxima, we need to evaluate the second derivative of $f(x)$ at each point.<br /><br />The second derivative of $f(x)$ is given by:<br />$f''(x) = \frac{d^2}{dx^2} \left( \frac{3}{5}x^5 + 12x^4 + 60x^3 + 3 \right)$<br />$f''(x) = 12x^3 + 144x^2 + 360x$<br /><br />Now, let's evaluate $f''(x)$ at each of the critical points:<br />$f''(0) = 0$<br />$f''(-10) = -1200$<br />$f''(-6) = -360$<br /><br />Since $f''(0) = 0$, we cannot determine whether $x = 0$ is a relative maximum or minimum. We need to use the first derivative test or the second derivative test to determine this.<br /><br />Using the first derivative test, we can see that the sign of $f'(x)$ changes from positive to negative as $x$ increases through $x = 0$. Therefore, $x = 0$ is a relative maximum.<br /><br />Using the second derivative test, we can see that $f''(-10) < 0$ and $f''(-6) < 0$. Therefore, $x = -10$ and $x = -6$ are relative minima.<br /><br />Therefore, the x-coordinate of the relative maximum of $f(x)$ is $x = 0$.