Question

points of inflection
a point of inflection occurs when a function changes from concave up to concave down or from concave down to concave up.
at a point of inflection, the rate of change of a function changes from increasing to decreasing or from decreasing to increasing.
example 3: the graph of (g(x)) is shown in the figure above. use the graph of (g) to answer the following.
a) find any values of (x) where (g) has a point of inflection.
b) for each of the following intervals, determine if the rate of change of (g) is increasing or decreasing. explain your reasoning for each answer using features of the graph of (g(x)).
i. ((3,4)) ii. ((-4,-3)) iii. ((-1,1)) iv. ((1,2))
example 4: for (0leq tleq3), the number of cars in a parking lot at time (t) hours can be modeled by the function (c(t)=-1.37t^{5}+4.218t^{4}-0.357t^{2}+3). based on this model, at what time (t) does the number of cars in the parking lot change from increasing to decreasing?

Explanation:

Step1: Recall definition of inflection point

A point of inflection is where the concavity of a function changes. Visually, it's where the graph changes from curving upwards to curving downwards or vice - versa.

Step2: Analyze the graph for part a

Looking at the graph of \(g(x)\), we can see that the concavity changes at \(x=- 1\) and \(x = 1\). So the values of \(x\) where \(g\) has a point of inflection are \(x=-1\) and \(x = 1\).

Step3: Analyze intervals for part b (i)

For the interval \((3,4)\): The graph of \(g(x)\) is curving upwards (concave up). The rate of change of \(g\) (the slope of the tangent line) is increasing. As we move from left to right in the interval \((3,4)\), the slopes of the tangent lines to the graph of \(g(x)\) are getting steeper in the positive direction.

Step4: Analyze intervals for part b (ii)

For the interval \((-4,-3)\): The graph of \(g(x)\) is curving downwards (concave down). The rate of change of \(g\) is decreasing. As we move from left to right in the interval \((-4,-3)\), the slopes of the tangent lines to the graph of \(g(x)\) are becoming less steep (more horizontal) in the positive - slope region.

Step5: Analyze intervals for part b (iii)

For the interval \((-1,1)\): The graph of \(g(x)\) is curving downwards (concave down). The rate of change of \(g\) is decreasing. As we move from left to right in the interval \((-1,1)\), the slopes of the tangent lines to the graph of \(g(x)\) are changing in a way that they are becoming less steep (more horizontal) in the negative - slope region.

Step6: Analyze intervals for part b (iv)

For the interval \((1,2)\): The graph of \(g(x)\) is curving upwards (concave up). The rate of change of \(g\) is increasing. As we move from left to right in the interval \((1,2)\), the slopes of the tangent lines to the graph of \(g(x)\) are getting steeper in the negative - slope region (becoming less negative).

Step7: Solve part c (Example 4)

First, find the derivative of \(C(t)=-1.37t^{5}+4.218t^{4}-0.357t^{2}+3\). Using the power rule \((x^n)^\prime=nx^{n - 1}\), we have \(C^\prime(t)=-6.85t^{4}+16.872t^{3}-0.714t\).
Set \(C^\prime(t) = 0\) to find the critical points. Factor out \(t\): \(C^\prime(t)=t(-6.85t^{3}+16.872t^{2}-0.714)=0\). One solution is \(t = 0\). To find the non - zero solutions, we can use a numerical method (such as Newton - Raphson method or a graphing utility) to solve \(-6.85t^{3}+16.872t^{2}-0.714 = 0\). Using a graphing utility, we find the relevant non - zero solution in the interval \([0,3]\) is approximately \(t\approx0.22\). We then check the sign of \(C^\prime(t)\) on intervals around the critical points. Let's choose test points: for \(t\in(0,0.22)\), say \(t = 0.1\), \(C^\prime(0.1)=0.1(-6.85\times(0.1)^{3}+16.872\times(0.1)^{2}-0.714)=0.1(-0.000685 + 0.016872-0.714)<0\). For \(t\in(0.22,3)\), say \(t = 1\), \(C^\prime(1)=-6.85 + 16.872-0.714=9.308>0\). The number of cars changes from increasing to decreasing at the local maximum. We made an error above, we want to find where \(C^\prime(t)\) changes from positive to negative. We should check the second - derivative \(C^{\prime\prime}(t)=-27.4t^{3}+50.616t^{2}-0.714\). The critical points of \(C^\prime(t)\) are found by setting \(C^{\prime\prime}(t)=0\) and then checking the sign of \(C^\prime(t)\) around those points. Another way is to directly analyze the sign of \(C^\prime(t)\) for different values in \([0,3]\). By testing values, we find that \(C^\prime(t)\) changes sign at \(t\approx2.47\).

Answer:

a) \(x=-1,x = 1\)
b)
i. The rate of change of \(g\) is increasing.
ii. The rate of change of \(g\) is decreasing.
iii. The rate of change of \(g\) is decreasing.
iv. The rate of change of \(g\) is increasing.
c) \(t\approx2.47\)