91Ó°ÊÓ

To begin, compute the first derivative of the given function using the power rule: $$f'(x) = \frac{d}{dx}(5x^4 - 20x^3 + 10) = 20x^3 - 60x^2$$

Next, compute the second derivative by differentiating f'(x) with respect to x, also using the power rule: $$f''(x) = \frac{d}{dx}(20x^3 - 60x^2) = 60x^2 - 120x$$

Now, set f''(x) equal to zero and solve for x to find the critical points of the second derivative (which correspond to potential inflection points of the original function): $$60x^2 - 120x = 0$$ Factor out the common factor of 60x: $$60x(x - 2) = 0$$ This equation is equal to zero when x = 0 or x = 2. These are the potentially inflection points.

To determine the intervals of concavity and identify inflection points, analyze the sign of f''(x) around our critical points (x = 0 and x = 2): 1. Choose a value less than 0, say x=-1; then f''(-1) = 60(-1)^2 - 120(-1) = 60 + 120 = 180 > 0, which means the function is concave up in the interval \((-\infty, 0)\). 2. Choose a value between 0 and 2, say x=1; then f''(1) = 60(1)^2 - 120(1) = 60 - 120 = -60 < 0, which means the function is concave down in the interval \((0, 2)\). 3. Choose a value greater than 2, say x=3; then f''(3) = 60(3)^2 - 120(3) = 180 > 0, which means the function is concave up in the interval \((2, \infty)\). As the concavity changes at both x = 0 and x = 2, these are inflection points. So, the function f(x) is concave up on the intervals \((-\infty, 0)\) and \((2, \infty)\), concave down on the interval \((0, 2)\), and has inflection points at x = 0 and x = 2.