Problem 68 - Single Variable Calculus

Question

Prove that \(x-x^{3} / 6<\sin x0\) Hint: To prove the left inequality, let \(f(x)=\sin x-x+x^{3} / 6\), and show that \(f\) is increasing on the interval \((0, \infty)\).

Step-by-Step Solution

Answer

To prove that \(x - \frac{x^3}{6} < \sin x < x\) for all \(x > 0\), we first analyze the function \(f(x) = \sin x - x + \frac{x^3}{6}\) and find its derivative \(f'(x) = \cos x - 1 + \frac{x^2}{2}\). We show that \(f'(x) > 0\) on the interval \((0, \infty)\) by analyzing the properties of the cosine and quadratic functions. Since \(f'(x) > 0\), it means that \(f(x)\) is increasing on the interval \((0, \infty)\), and thus \(f(x) > 0\) for all \(x > 0\), proving the left inequality. For the right inequality, we use the properties of the sine function to show that \(\sin x < x\) for all \(x > 0\). By combining the results, we can conclude that for all \(x > 0\), \(x - \frac{x^3}{6} < \sin x < x\).

Step 1: Define the given function and find its derivative \((f'(x))\)

Let \(f(x) = \sin x - x + \frac{x^3}{6}\). Now let's find the derivative of \(f(x)\) with respect to \(x\). Using the rules of differentiation, we have: \[f'(x) = \cos x - 1 + \frac{3x^2}{6}\]

Step 2: Determine where the derivative is positive

Now, we simplify the expression for \(f'(x)\): \[f'(x) = \cos x - 1 + \frac{x^2}{2}\] Our goal is to show that the derivative, \(f'(x)\), is positive on the interval \((0, \infty)\). To do this, we can look for critical points where the derivative is 0 or undefined: \[f'(x) = \cos x - 1 + \frac{x^2}{2} = 0\] There are no obvious solutions to this equation in closed form. However, we can use a graphical approach to show that the properties of the cosine and quadratic functions ensure that the whole expression remains positive on the interval \((0, \infty)\).

Step 3: Analyze properties of the cosine and quadratic function

Recall that the range of the cosine function is \([-1, 1]\). On the interval \((0, \infty)\), \(\cos(x) - 1 \leq 0\), since the cosine function is always weakly decreasing on this interval (note that \(\cos(\pi)=-1\), hence \(\cos(x) \leq 1\).) Additionally, the quadratic function \(\frac{x^2}{2}\) is positive for all \(x > 0\). This means that the sum of these two functions, \(f'(x)\), is always positive for \(x > 0\).

Step 4: Prove the left inequality

Since \(f'(x) > 0\) on the interval \((0,\infty)\), we know that the function \(f(x) = \sin x - x + \frac{x^3}{6}\) is increasing on the interval \((0, \infty)\). Since \(f(0) = 0\), we can conclude that \(f(x) > 0\) for all \(x > 0\), which implies that \[\sin x - x + \frac{x^3}{6} > 0\] Therefore, the left inequality is proved: \[x - \frac{x^3}{6} < \sin x\]

Step 5: Prove the right inequality

Now we need to prove the right inequality \(\sin x < x\). Notice that the sine function is bound by \(-1 \leq \sin x \leq 1\), and for \(x > 0\), the sine function is positive, and less than its arclength, so \(\sin x < x\). Therefore, the right inequality is proved: \[\sin x < x\]

Step 6: Combine the left and right inequality to derive the final result

Now that we have proved both the left inequality, \(x - x^3/6 < \sin x\) and the right inequality, \(\sin x < x\), we can conclude that for all \(x > 0\), the following inequality holds: \[x - \frac{x^3}{6} < \sin x < x\]