Chapter 3: Problem 60
Find the minimum value or maximum value of the function. Then describe where the function is increasing and decreasing. (Section 2.2) \(h(x)=x^2+3 x-18\)
Short Answer
Expert verified
The function \(h(x)=x^2+3 x-18\) has a minimum value of -20.25 at \(x = -1.5\). The function is decreasing for \(x < -1.5\) and increasing for \(x > -1.5\).
Step by step solution
01
Find the Derivative
The derivative of the function \(h(x)\) can be found using the power rule for differentiation, which states that the derivative of \(x^n\) is \(nx^{n-1}\). Thus, the derivative of \(h(x)=x^2+3 x-18\) is \(h'(x) = 2x+3\).
02
Find the Critical Points
The critical points occur where the derivative equals zero or does not exist. Hence, to find the critical points solve the equation \(h'(x) = 2x + 3 = 0\). Solving this equation gives \(x = -3/2 = -1.5\).
03
Determine where the function is increasing or decreasing
To classify the intervals of increase and decrease, we'll use a number line and the critical points. The function will decrease to the left of \(x = -1.5\), since \(h'(-1.6) = 2(-1.6)+3 < 0\) and it will increase to the right of \(x = -1.5\), since \(h'(-1.4) = 2(-1.4) + 3 > 0\).
04
Find the maximum or minimum value of the function
As we see from Step 3, the function decreases before \(x = -1.5\) and increases after. Hence, at \(x = -1.5\), the function reaches a minimum. In order to find out the minimum value of the function we substitute \(x = -1.5\) in the original function \(h(x)=x^2+3 x-18\). So, the minimum value of the function is \(h(-1.5) = (-1.5)^2 +3(-1.5) - 18 = -20.25\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differentiation
Differentiation is a fundamental concept in calculus that helps us understand how a function changes at any given point. When we differentiate a function, we find its derivative, which gives us the rate of change. For polynomial functions like quadratics, we apply the power rule. This rule states that the derivative of a function of the form \(x^n\) is \(nx^{n-1}\).
In the original problem, we're dealing with the quadratic function \(h(x) = x^2 + 3x - 18\). By applying the power rule here, we find the derivative to be \(h'(x)=2x+3\).
In the original problem, we're dealing with the quadratic function \(h(x) = x^2 + 3x - 18\). By applying the power rule here, we find the derivative to be \(h'(x)=2x+3\).
- "2x" is derived from \(x^2\) by multiplying the exponent 2 by \(x\) and reducing the exponent by one.
- The derivative of a constant, like \(-18\), is zero.
Critical Points
Critical points of a function occur where its derivative is zero or undefined. These points are important as they may signify local maxima or minima, or points of inflection.
In our function \(h(x)=x^2+3x-18\), we set its derivative \(h'(x)=2x+3\) to zero in order to find the critical points:
\(2x+3=0\).
Solving for \(x\), we find the critical point at \(x=-1.5\).
Now that we have found \(x=-1.5\) as a critical point, we can use it to further investigate the function's behavior around this point. This involves checking where the function increases or decreases and whether this point is a minimum or a maximum.
In our function \(h(x)=x^2+3x-18\), we set its derivative \(h'(x)=2x+3\) to zero in order to find the critical points:
\(2x+3=0\).
Solving for \(x\), we find the critical point at \(x=-1.5\).
Now that we have found \(x=-1.5\) as a critical point, we can use it to further investigate the function's behavior around this point. This involves checking where the function increases or decreases and whether this point is a minimum or a maximum.
Increasing and Decreasing Intervals
To determine where a function is increasing or decreasing, we examine the sign of its derivative around critical points.
For the quadratic function \(h(x)=x^2+3x-18\), we identified \(x = -1.5\) as a critical point. Now let's perform a test by selecting points on either side of \(x = -1.5\) and calculating their derivative values:
For the quadratic function \(h(x)=x^2+3x-18\), we identified \(x = -1.5\) as a critical point. Now let's perform a test by selecting points on either side of \(x = -1.5\) and calculating their derivative values:
- Choose a number less than \(-1.5\), say \(-1.6\). Calculate \(h'(-1.6) = 2(-1.6)+3 = -0.2\), which is negative, indicating the function is decreasing here.
- Choose a number greater than \(-1.5\), say \(-1.4\). Calculate \(h'(-1.4) = 2(-1.4)+3 = 0.2\), which is positive, indicating the function is increasing here.
Minimum and Maximum Values
The minimum or maximum values of a function are determined by analyzing its critical points and behavior in intervals.
For \(h(x) = x^2 + 3x - 18\), we found the critical point at \(x = -1.5\). From the examination of increasing and decreasing intervals, we saw that the function decreases before this point and increases after. This suggests that \(x=-1.5\) is a minimum point.
To find the actual minimum value at this point, substitute \(x = -1.5\) back into the original function: \(h(-1.5) = (-1.5)^2 + 3(-1.5) - 18\).
Calculating, we get \(h(-1.5) = -20.25\). Thus, \(-20.25\) is the minimum value of the function and there is no maximum because a quadratic with a positive leading function like this one opens upwards.
For \(h(x) = x^2 + 3x - 18\), we found the critical point at \(x = -1.5\). From the examination of increasing and decreasing intervals, we saw that the function decreases before this point and increases after. This suggests that \(x=-1.5\) is a minimum point.
To find the actual minimum value at this point, substitute \(x = -1.5\) back into the original function: \(h(-1.5) = (-1.5)^2 + 3(-1.5) - 18\).
Calculating, we get \(h(-1.5) = -20.25\). Thus, \(-20.25\) is the minimum value of the function and there is no maximum because a quadratic with a positive leading function like this one opens upwards.