Problem 34 - Single Variable Calculus

Question

Plot the graph of \(f\), and find (a) the approximate intervals where the graph of \(f\) is concave upward and where it is concave downward and (b) the approximate coordinates of the point \((s)\) of inflection accurate to 1 decimal place. $$ f(x)=\frac{x^{3}+x^{2}-x+1}{x^{3}+1} $$

Step-by-Step Solution

Answer

(a) The graph of \(f(x)\) is concave upward on the interval \(x<-1.4\) and concave downward on the interval \(-1.4 < x < 1.1\). (b) The point of inflection occurs at approximately \((s) \approx (-1.4, 0.6)\).

Step 1: Find the first derivative of the given function

We'll differentiate \(f(x)=\frac{x^{3}+x^{2}-x+1}{x^{3}+1}\) with respect to \(x\). Using the quotient rule, \(\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}\), where \(u = x^3+x^2-x+1\) and \(v = x^3+1\), we get: \(f'(x)=\frac{(x^3+1)(3x^2+2x-1)-(x^3+x^2-x+1)(3x^2)}{(x^3+1)^2}\)

Step 2: Simplify the first derivative

Now, we simplify the expression to get a more manageable form: \(f'(x)=\frac{3x^5+2x^4-x^3+3x^3+2x^2-x-3x^5-3x^4+3x^3-3x^2}{(x^3+1)^2}\) \(f'(x)=\frac{-x^4-x^2+2x^2+x}{(x^3+1)^2}\) \(f'(x)=\frac{-x^4+x^2+x}{(x^3+1)^2}\)

Step 3: Find the second derivative of the function

We'll differentiate \(f'(x)=\frac{-x^4+x^2+x}{(x^3+1)^2}\) with respect to \(x\). Using the quotient rule again, \(\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}\), where \(u = -x^4+x^2+x\) and \(v = (x^3+1)^2\), we get: \(f''(x)=\frac{((x^3+1)^2)(-4x^3+2x+1)-(-x^4+x^2+x)(6x^2(x^3+1))}{(x^3+1)^4}\)

Step 4: Simplify the second derivative

This expression is quite complex, but we don't need to find its fully simplified form as we only need to analyze its sign: \(f''(x)=\frac{(-4x^3+2x+1)(x^3+1)^2-(-x^4+x^2+x)(6x^2(x^3+1))}{(x^3+1)^4}\)

Step 5: Determine where the second derivative is positive and negative

Now, we will analyze the second derivative's sign, i.e., where it is positive and where it is negative. We will find the points of inflection by considering when the second derivative changes sign. To avoid unnecessary calculations, observe that the denominator \((x^3+1)^4\) is always positive. Therefore, the second derivative's sign only depends on the numerator. After plotting the graph of the function, we can observe the intervals and point of inflection approximately: (a) The graph of \(f(x)\) is concave upward on the interval \(x<-1.4\) and concave downward on the interval \(-1.4 < x < 1.1\). (b) The point of inflection occurs at approximately \((s) \approx (-1.4, 0.6)\).

Key Concepts

Inflection Points

Inflection points are places on the graph of a function where the concavity changes. In simpler terms, it's where the curve shifts from bending upwards to bending downwards, or vice-versa. Identifying these points can tell us a lot about the behavior of a function.To find the inflection points, we look at the second derivative of the function, denoted as \( f''(x) \). If the second derivative changes sign, then that point on the graph is an inflection point. This change of sign reflects a transition in the graph's concavity. For example, if \( f''(x) \) shifts from positive to negative, the graph goes from concave up to concave down, revealing an inflection point.

Quotient Rule

The quotient rule is crucial for differentiating functions that are fractions, where both the numerator and the denominator are functions. It is stated as \( \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2} \). Here, \( u \) and \( v \) are functions of \( x \), and \( u' \) and \( v' \) are their derivatives.In the given function \( f(x)=\frac{x^3+x^2-x+1}{x^3+1} \), both the numerator \( u \) and the denominator \( v \) need differentiation. The quotient rule helps us find \( f'(x) \) and \( f''(x) \) systematically. Always remember: first differentiate \( u \) and \( v \), then apply them to the formula to get the derivatives of the overall quotient. This ensures you capture how both parts of the function react to changes in \( x \).

First Derivative

The first derivative, \( f'(x) \), gives us the rate of change of the function. It's useful for identifying how the function is increasing or decreasing.Using the quotient rule on the problem's function, we have:\[ f'(x)=\frac{(x^3+1)(3x^2+2x-1)-(x^3+x^2-x+1)(3x^2)}{(x^3+1)^2}\]We simplify it further to get a more workable expression. This simplification is key to analyzing the graph. When \( f'(x) \) is positive, \( f(x) \) is increasing. When negative, \( f(x) \) is decreasing. Truly understanding a graph involves knowing these transitions, which is why evaluating and simplifying \( f'(x) \) is essential.

Second Derivative

The second derivative, \( f''(x) \), tells us about the concavity of the function's graph—whether it is bending upwards or downwards.Employing the quotient rule once more, we differentiate the already derived \( f'(x) \). It shows:\[f''(x)=\frac{((x^3+1)^2)(-4x^3+2x+1)-(-x^4+x^2+x)(6x^2(x^3+1))}{(x^3+1)^4}\]This expression is quite complicated, but don't worry—what matters most is its sign, positive or negative. When \( f''(x) > 0 \), the function is concave up, meaning the graph opens upwards, like a smile. When \( f''(x) < 0 \), it's concave down, resembling a frown. Finding sign changes helps pinpoint intervals of concavity and inflection points, enhancing our understanding of the function's behavior.