91Ó°ÊÓ

To find where the function is concave upward or downward, we need to determine the second derivative of the function. First, let's find the first derivative of the function, denoted as \(f'(t)\). We will use the power rule of differentiation: \(\frac{d}{dx}x^n = nx^{n-1}\) $$ f'(t) = \frac{d}{dt}(t^4 - 2t^3) = 4t^3 - 6t^2 $$

Now that we've found the first derivative, let's find the second derivative, denoted as \(f''(t)\). Again, we will use the power rule of differentiation. $$ f''(t) = \frac{d}{dt}(4t^3 - 6t^2) = 12t^2 - 12t = 12t(t - 1) $$

To find the inflection points and intervals of concavity, we need to find the critical points of the second derivative, which are the points where \(f''(t) = 0\) or \(f''(t)\) is undefined. In this case, \(f''(t)\) is never undefined, so we only need to solve for when \(f''(t) = 0\). $$ 12t(t - 1) = 0 $$ This equation has two solutions: \(t = 0\) and \(t = 1\). These are the critical points of the second derivative.

Now that we have the critical points of the second derivative, we can use them to determine the intervals of concavity and the inflection points. For the intervals of concavity, we will test the second derivative using values in the intervals created by the critical points. 1. Test the interval \((-∞, 0)\): Choose a value, say -1, and plug it into \(f''(t)\): \(f''(-1) = -12 < 0\), so this interval is concave downward. 2. Test the interval \((0, 1)\): Choose a value, say 0.5, and plug it into \(f''(t)\): \(f''(0.5) = 3 > 0\), so this interval is concave upward. 3. Test the interval \((1, ∞)\): Choose a value, say 2, and plug it into \(f''(t)\): \(f''(2) = 24 > 0\), so this interval is concave upward. Now we can summarize our findings: Concave upward: \(0 < t < 1\) and \(t > 1\) Concave downward: \(t < 0\) For inflection points, they occur where the concavity changes. From our analysis above, we can see that the concavity changes at the critical points \(t = 0\) and \(t = 1\). Therefore, the inflection points are at \(t = 0\) and \(t = 1\). To find the corresponding function values, we plug these points back into the original function: Inflection point 1: \((0, f(0)) = (0, 0)\) Inflection point 2: \((1, f(1)) = (1, -1)\) So, we have found that the graph of the function is concave upward when \(0 < t < 1\) and \(t > 1\). It is concave downward when \(t < 0\). The inflection points are at \((0, 0)\) and \((1, -1)\).