Answer
a.) g'(0) = -3
b.) g'(3) = 0
c.) Since g'(x) < 0 at x = 1, g(x) is DECREASING at x = 1 at a rate of -$\dfrac{8}{3}$
d.) Since g'(x) > 0 at x = -4, g(x) is INCREASING at x = -4 at a rate of $\dfrac{7}{3} $
e.) g(6) - g(4) is POSITIVE because g'(x) is positive and increasing on the interval $4 \leq x \leq 6$. This means g(x) is concave up and increasing on this interval and therefore g(6) > g(4). See explanation for more details.
f.) No, the function g'(x) has not been given.
Work Step by Step
a.) By looking at the graph, you can see that $g'(0) = -3$ since the g'(x) is in between y = -2 and y = -4 at x = 0
b.) By looking at the graph, you can see that $g'(3) = 0$ since g'(x) crosses the x-axis at x = 3
c and d.) The derivative can tell us the slope of a function at a certain x-value. Knowing the g'(1) is less than zero and g'(-4) is greater than zero tells us which way the graph is heading at that point. When the derivative (g'(x)) is negative, it tells us that g(x) is decreasing since the slope at that point is negative (this means that the line tangent to the graph at that point is approaching negative infinity as x increases). The opposite is also true. If g'(x) is positive, it means that g(x) is increasing. This is the first derivative test that you will learn more about in chapter 3, "Applications of Derivatives"
e.) This question is easiest to answer in terms of the second derivative test. First, we can conclude that g(x) is increasing between x = 4 and x = 6 since g'(x) is positive on this interval ($4 \leq x \leq 6$). This does not mean that both g(6) and g(4) are positive. While we can't find exact values for g(6) and g(4), the graph can still tell us that g(6) > g(4). Why? Because g'(x) is increasing on this interval. That means that the slope is increasing as x is increasing, so g(x) is concave up.
f.) You can find the exact value of a point on g(x) if you are given the function of g'(x), but you are not.
