Interpreting Graphs
Contents
Example
Question
Consider the graph of the derivative of \(g(x)\).
- For which values of \(x\) is \(g(x)\) decreasing?
- Determine the \(x\)-coordinate(s) of the turning point(s) of \(g(x)\).
- Given that \(g(x) = ax^{3} + bx^{2} + cx\), calculate \(a\), \(b\) and \(c\).
Examine the parabolic graph and interpret the given information
We know that \(g'(x)\) describes the gradient of \(g(x)\). To determine where the cubic function is decreasing, we must find the values of \(x\) for which \(g'(x) < 0\):
\[\{x: – 2 < x < 1; x \in \mathbb{R} \} \text{ or we can write } x \in (-2;1)\]
Determine the \(x\)-coordinate(s) of the turning point(s)
To determine the turning points of a cubic function, we let \(g'(x) = 0\) and solve for the \(x\)-values. These \(x\)-values are the \(x\)-intercepts of the parabola and are indicated on the given graph:
\[x = -2 \text{ or } x = 1\]
Determine the equation of \(g(x)\)
\begin{align*} g(x) &= ax^{3} + bx^{2} + cx \\ g'(x) &= 3ax^{2} + 2bx + c \end{align*}
From the graph, we see that the \(y\)-intercept of \(g'(x)\) is \(-\text{6}\).
\begin{align*} \therefore c &= – 6 \\ g'(x) &= 3ax^{2} + 2bx – 6 \\ \text{Substitute } x = -2: \enspace g'(-2) &= 3a(-2)^{2} + 2b(-2) – 6 \\ 0 &= 12a – 4b – 6 \ldots \ldots (1) \\ \text{Substitute } x = 1: \enspace g'(1) &= 3a(1)^{2} + 2b(1) – 6 \\ 0 &= 3a + 2b – 6 \ldots \ldots (2) \end{align*}\begin{align*} \text{Eqn. }(1) – 4 \text{ Eqn. }(2): \quad 0 &= 0 -12b + 18 \\ \therefore b &= \frac{3}{2} \\ \text{And } 0 &= 3a + 2 \left( \frac{3}{2} \right) – 6 \\ 0 &= 3a – 3 \\ \therefore a &= 1 \\ & \\ g(x) &= x^{3} + \frac{3}{2}x^{2} – 6x \end{align*}
