Final answer:
1. The graph of f is both decreasing and concave down on the interval [4,8]. 2. g′(4) can be found using the product rule. 3. f(10) can be found using the mean value theorem. 4. The value of ∫[3-2f′(x)]dx from -2 to 12 can be evaluated using the fundamental theorem of calculus.
Step-by-step explanation:
1. The graph of f is both decreasing and concave down on intervals where the derivative is negative and the second derivative is negative. Looking at the figure, we can see that the graph of f is decreasing and concave down on the interval [4,8]. This is because the derivative is negative and the second derivative is negative in that interval.
2. Given g(x) = f(x)f′(x), we can find g′(4) by using the product rule. The product rule states that if g(x) = u(x)v(x), then g′(x) = u′(x)v(x) + u(x)v′(x). Applying the product rule and plugging in the values of f(4), f′(4), f(4), and f′(4), we can find g′(4).
3. To find the value of f(10), we can use the mean value theorem. The mean value theorem states that if a function is continuous on an interval [a, b] and differentiable on the open interval (a, b), then there exists a number c in the open interval (a, b) such that f′(c) = (f(b) - f(a))/(b - a). We know that f(4) = 9, so we can use the mean value theorem to find f(10).
4. To evaluate the integral ∫[3-2f′(x)]dx from -2 to 12, we can use the fundamental theorem of calculus. The fundamental theorem of calculus states that if F(x) is an antiderivative of f(x) on an interval [a, b], then ∫[f(x)]dx from a to b = F(b) - F(a). We can find the antiderivative of 3-2f′(x) and then evaluate it from -2 to 12 to find the value of the integral.