91Ó°ÊÓ

To find the intervals of concavity, we need to find the first derivative of the function, g'(t). Using the power rule, we have: $$g'(t) = \frac{d}{dt}(3 t^{5}-30 t^{4}+80 t^{3}+100) = 15 t^{4} - 120 t^{3} + 240 t^{2}$$

Now, we will find the second derivative, g''(t), which will help us determine the intervals of concavity: $$g''(t) = \frac{d^2}{dt^2}(15 t^{4} - 120 t^{3} + 240 t^{2}) = 60 t^{3} - 360 t^{2} + 480 t$$

To find the critical points of g''(t), we set g"(t) equal to zero and solve for t: $$60 t^{3} - 360 t^{2} + 480 t = 0$$ We can notice that all terms have a common factor of 60, so we can simplify the equation by dividing both sides by 60: $$t^3 - 6t^2 + 8t = 0$$ Now, we can factor out a t term: $$t (t^2 - 6t + 8) = 0$$

Now, we will solve the factored equation for t: For t = 0: $$t = 0$$ For the quadratic equation, t^2 - 6t + 8 = 0, we can factor it to $$(t - 2)(t - 4) = 0$$ Thus, we have the critical points t = 0, 2, and 4.

We will use a number line and our critical points to find where the function is concave up or down: - Take the interval (-∞, 0), we can choose any test point, for example, t = -1. $$g''(-1)=60(-1)^3 - 360(-1)^2 + 480(-1)=-60-360-480$$which is negative, so the function is concave down in this interval. - For the interval (0, 2), choose t = 1: $$g''(1)=60(1)^3 - 360(1)^2 + 480(1)=60-360+480$$which is positive, so the function is concave up in this interval. - For the interval (2, 4), choose t = 3: $$g''(3)=60(3)^3 - 360(3)^2 + 480(3)=540-3240+1440$$ which is negative, so the function is concave down in this interval. - For the interval (4, ∞), choose t = 5: $$g''(5)=60(5)^3 - 360(5)^2 + 480(5)=3000-9000+2400$$which is positive, so the function is concave up in this interval. So the function is: - Concave down on the intervals (-∞, 0) and (2, 4), - Concave up on the intervals (0, 2) and (4, ∞), - Inflection points are at t = 0, 2, and 4, which are the points where concavity changes. The corresponding points on the function are g(0)=100, g(2)=100, and g(4)=100.