Question

Find the critical points of the function (i) \(f(x)=\frac{2}{3} \sqrt{x^{3}}-\frac{x}{2}+\int_{1}^{x}\left(\frac{1}{2}+\frac{1}{2} \cos 2 t-t^{\frac{1}{2}}\right) d\) (ii) \(f(x)=\int_{0}^{x}\left(\sin ^{2} 2 t-2 \cos ^{2} 2 t+a\right) d t\)

Short answer

Question: Find the critical points of the following functions: (i) \(f(x)=\frac{2}{3}\sqrt{x^{3}}-\frac{x}{2}+\int_{1}^{x}\left(\frac{1}{2}+\frac{1}{2} \cos 2t-t^{\frac{1}{2}}\right)dt\) (ii) \(f(x) = \int_{0}^{x}\left(\sin^{2}(2t)-2\cos^{2}(2t) + a\right) dt\) Answer: (i) The critical points of \(f(x)\) occur at \(x = \frac{(2n + 1)\pi}{4}\), where n is an integer. (ii) The critical points of \(f(x)\) depend on the value of the parameter a. Further information about the value of a is needed to find an explicit solution for x.

Step-by-step solution

Problem (i) - Differentiate the function

To find the critical points for the function \(f(x)=\frac{2}{3}\sqrt{x^{3}}-\frac{x}{2}+\int_{1}^{x}\left(\frac{1}{2}+\frac{1}{2} \cos 2t-t^{\frac{1}{2}}\right)dt\), we need to differentiate it with respect to x using the Fundamental Theorem of Calculus for the integral term.

Problem (i) - Fundamental Theorem of Calculus usage

The derivative of the integral term \(\int_{1}^{x}\left(\frac{1}{2}+\frac{1}{2} \cos 2t-t^{\frac{1}{2}}\right) dt\) with respect to x is: \(\frac{1}{2}+\frac{1}{2} \cos(2x)-x^{\frac{1}{2}}\).

Problem (i) - Differentiate the non-integral terms

Next, differentiate the non-integral terms of the function. The derivative of \(\frac{2}{3}\sqrt{x^{3}}\) with respect to x is \(\frac{2}{3} \times \frac{3}{2}x^{\frac{3}{2}-1} = x^{\frac{1}{2}}\). The derivative of \(-\frac{x}{2}\) with respect to x is \(-\frac{1}{2}\).

Problem (i) - Combine all derivative terms

Now, we combine all derivative terms to find the derivative of \(f(x)\) with respect to x: \(f'(x) = x^{\frac{1}{2}} - \frac{1}{2} + \frac{1}{2} + \frac{1}{2} \cos(2x) - x^{\frac{1}{2}} = \frac{1}{2} \cos(2x)\).

Problem (i) - Find the critical points

To find the critical points, we set \(f'(x)\) equal to zero: \(0 = \frac{1}{2} \cos(2x)\). Therefore, \(\cos(2x) = 0\). The solution of this equation is: \(x = \frac{(2n + 1)\pi}{4}\), where n is an integer.

Problem (ii) - Differentiate the function

To find the critical points for the function \(f(x) = \int_{0}^{x}\left(\sin^{2}(2t)-2\cos^{2}(2t) + a\right) dt\) we need to differentiate it with respect to x using the Fundamental Theorem of Calculus for the integral term.

Problem (ii) - Fundamental Theorem of Calculus usage

The derivative of the integral term \(\int_{0}^{x}\left(\sin^{2}(2t)-2\cos^{2}(2t)+a\right) dt\) with respect to x is: \(\sin^{2}(2x) - 2\cos^{2}(2x) + a\)

Problem (ii) - Find the critical points

To find the critical points, we set \(f'(x)\) equal to zero: \(0 = \sin^{2}(2x) - 2\cos^{2}(2x) + a\). This equation involves the parameter a and trigonometric functions. Without information about the value of a, we cannot find an explicit solution for x. If further information is provided about a, we can proceed with solving for the critical points.