Find the x -coordinates of all relative maxima of f(x) f(x)=(3)/(5)x^5+12x^4+60x^3+3

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 the function.<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 are:<br />$x = 0, x = -10, x = -6$<br /><br />Step 3: Determine which of the critical points correspond to relative maxima.<br />To determine which of the 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 />$f''(x) = \frac{d^2}{dx^2} \left( \frac{3}{5}x^5 + 12x^4 + 60x^3 + 3 \right)$<br />$f''(x) = 12x^3 + 144x^2 + 360x$<br /><br />Now, let's evaluate the second derivative at each critical point:<br /><br />For $x = 0$:<br />$f''(0) = 12(0)^3 + 144(0)^2 + 360(0) = 0$<br /><br />For $x = -10$:<br />$f''(-10) = 12(-10)^3 + 144(-10)^2 + 360(-10) = -1200 + 14400 - 3600 = 10800 > 0$<br /><br />For $x = -6$:<br />$f''(-6) = 12(-6)^3 + 144(-6)^2 + 360(-6) = -2592 + 5184 - 2160 = 432 > 0$<br /><br />Since the second derivative is positive at $x = -10$ and $x = -6$, these points correspond to relative minima. Therefore, there are no relative maxima for the given function.<br /><br />Therefore, the x-coordinates of all relative maxima of $f(x) = \frac{3}{5}x^5 + 12x^4 + 60x^3 + 3$ are $\boxed{\text{None}}$.