Chapter 24: Problem 23
Sketch the graphs of the given functions by determining the appropriate information and points from the first and second derivatives. Use a calculator to check the graph. In Exercises \(27-32,\) use the function maximum-minimum feature to check the local maximum and minimum points. $$y=2 x^{3}+6 x^{2}-5$$
Short Answer
Step by step solution
Find the First Derivative
Solve for Critical Points
Find the Second Derivative
Determine Concavity and Inflection Points
Identify Local Maximum and Minimum
Sketch the Graph and Validate with Calculator
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First Derivative
This expression helps us find the critical points, which are values where the function's slope is zero, signaling potential maxima, minima, or points of inflection. Moreover, the sign of the first derivative informs us about the intervals on which the function is increasing or decreasing:
- If \(\frac{dy}{dx} > 0\), the function is increasing;
- If \(\frac{dy}{dx} < 0\), the function is decreasing.
Second Derivative
The second derivative can be used to determine:
- If \(\frac{d^2y}{dx^2} > 0\), the function is concave up ("cup" shape), indicating that the slope of the tangent is increasing.
- If \(\frac{d^2y}{dx^2} < 0\), the function is concave down ("cap" shape), indicating that the slope is decreasing.
Critical Points
For our function \(6x^2 + 12x\), setting it equal to zero gives the equation \(6x(x + 2) = 0\). Solving this, we get \(x = 0\) and \(x = -2\). These values are the critical points, which are then evaluated to classify them as either local maxima, local minima, or saddle points. Such evaluations involve substituting back into either the second derivative or the original function. Per our findings:
- At \(x = -2\), the point is a local maximum because the second derivative is negative;
- At \(x = 0\), the point is a local minimum as the second derivative is positive.
Concavity
- When \(\frac{d^2y}{dx^2}\) is greater than zero, the graph curves upwards;
- If less than zero, the graph bends downwards.
Inflection Points
In the example function, when evaluating \(12x + 12\), setting it to zero gives the potential for an inflection point. However, it's essential to confirm that the concavity actually changes by checking intervals around the point. Unfortunately, for our specific function evaluation, there are no inflection points that cause the concavity change between the critical points \(x = 0\) and \(x = -2\). This suggests that between these points, the function doesn't switch between curving up and down, maintaining continuous concavity.