Question

Graph the function. Identify the \(x\)-intercepts and the points where the local maximums and local minimums occur. Determine the intervals for which the function is increasing or decreasing. $$g(x)=2 x^3+8 x^2-3$$

Short answer

The solution involves mainly understanding and applying concepts from calculus, such as differentiation, setting the expression equal to zero, as well as the second derivative test. Its final results will be determined by the specific algebraic calculations and sign tests. For a more precise step by step solution, solving algebraically is necessary.

Step-by-step solution

Find the X-Intercepts

Set \(g(x) = 0\), which gives \(2x^3 + 8x^2 - 3 = 0\). Solve this equation to find the x-intercepts of the function.

Find Local Maxima and Minima

Consider \(g'(x)\), the first derivative of \(g(x)\), which is \(6x^2 + 16x\). Set this derivative equal to zero, and solve for \(x\). The values of \(x\) obtained are the critical points of the function. To classify these as local maximums or minimums, substitute these values back into the second derivative of the function and check their signs; if the second derivative at that point is positive, then \(g(x)\) has a local minimum at that \(x\), and if it is negative, then \(g(x)\) has a local maximum at that \(x\).

Determine Increasing and Decreasing Intervals

Again, consider \(g'(x)\). Where \(g'(x) > 0\), the function is increasing, and where \(g'(x) < 0\), the function is decreasing. Plug in some test values in the intervals between the roots of \(g'(x)\), and use their results to establish these intervals.

Key concepts

Finding X-Intercepts

To understand where a polynomial function like g(x) = 2x^3 + 8x^2 - 3 crosses the x-axis, we need to find its x-intercepts.

The x-intercepts, also known as zeros or roots, are the points where the graph of the function touches or crosses the x-axis. To find these, we set the function equal to zero and solve for x. For our function, setting g(x) = 0 leads to a cubic equation 2x^3 + 8x^2 - 3 = 0. Finding solutions to this equation can sometimes be done by factoring, but often requires numerical methods or graphing technology for more complicated polynomials. These solutions represent the x-values where the function's graph intersects the x-axis.

Local Maxima and Minima

The points on the graph of a function where it reaches a peak or a valley are known as local maxima and minima, respectively.

These are the points where the function changes direction from increasing to decreasing (local maxima) or from decreasing to increasing (local minima). To find these, we calculate the first derivative of the function, g'(x), which reveals the function's slope at any given point. By setting g'(x) = 0, we identify what are known as the critical points. However, not all critical points are local maxima or minima, which is why we use the second derivative test (discussed in another section) to determine the nature of these points.

Increasing and Decreasing Intervals

Determining where a function is increasing or decreasing is crucial for understanding its overall shape.

We use the first derivative g'(x) to find these intervals. If g'(x) > 0 for a particular interval, the function is increasing on that interval. Conversely, if g'(x) < 0, the function is decreasing. By testing values between the critical points in the derivative, we can establish these intervals. It's important to note that where the first derivative is zero, the function may have a horizontal tangent, indicating a potential local maxima or minima.

First Derivative Test

The First Derivative Test is a method used to determine if a critical point of a function is a local maximum, local minimum, or neither. It involves looking at the sign of the first derivative g'(x) to the left and right of the critical point.

If g'(x) changes from positive to negative as we pass through the critical point, the function has a local maximum there. If g'(x) changes from negative to positive, it has a local minimum. However, if g'(x) does not change signs, the test is inconclusive, and the critical point may not be a local extremum.

Second Derivative Test

The Second Derivative Test provides another way to classify critical points. After identifying the critical points using the first derivative, we take the second derivative of the function, g''(x), which measures the curvature of the graph.

By substituting the critical points into the second derivative, we can determine the concavity at those points. If g''(x) > 0, the graph is concave up, indicating a local minimum. If g''(x) < 0, the graph is concave down, and we have a local maximum. If g''(x) = 0, the test is inconclusive, and we may need to resort to further analysis to classify the point.