Free Math Worksheets

- What derivative value corresponds to a horizontal tangent? - Use both the product rule and the chain rule. - Include the derivative of the inner expression \(c-x\). - Substitute \(x=\frac{c}{2}\) before doing unnecessary simplification.

1. Differentiate using the product and chain rules: \(f'(x)=\sec^2(x)\tan(c-x)-\tan(x)\sec^2(c-x)\). 2. Substitute \(x=\frac{c}{2}\): \(f'\left(\frac{c}{2}\right)=\sec^2\left(\frac{c}{2}\right)\tan\left(\frac{c}{2}\right)-\tan\left(\frac{c}{2}\right)\sec^2\left(\frac{c}{2}\right)=0\). 3. Therefore, the tangent slope is \(0\), so the graph has a horizontal tangent at \(x=\frac{c}{2}\). The condition that \(f\) is defined there is equivalent to \(\cos\left(\frac{c}{2}\right)\neq0\).

\(f'\left(\frac{c}{2}\right)=0\), so the tangent is horizontal whenever \(\cos\left(\frac{c}{2}\right)\neq0\).

- Determine the values of cosine on the stated interval. - Use point-slope form for the tangent line. - Find both axis intercepts of the tangent line. - Use the area formula for a right triangle.

1. On the given interval, \(0\leq\cos(x)\leq1\). Since \(k>0\), the range is \([0, k]\). 2. Since \(h_k\left(\frac{\pi}{4}\right)=\frac{k\sqrt{2}}{2}\) and \(h_k'(x)=-k\sin(x)\), the tangent slope is \(-\frac{k\sqrt{2}}{2}\). Thus \(t_k: y=-\frac{k\sqrt{2}}{2}\left(x-\frac{\pi}{4}\right)+\frac{k\sqrt{2}}{2}\). 3. The y-intercept is \(\frac{k\sqrt{2}}{2}\left(1+\frac{\pi}{4}\right)\), and the x-intercept is \(1+\frac{\pi}{4}\). Therefore, \(A(k)=\frac{1}{2}\left(1+\frac{\pi}{4}\right)\frac{k\sqrt{2}}{2}\left(1+\frac{\pi}{4}\right)=\frac{k\sqrt{2}}{4}\left(1+\frac{\pi}{4}\right)^2\). 4. Set \(A(k)=10\): \(\frac{k\sqrt{2}}{4}\left(1+\frac{\pi}{4}\right)^2=10\). Solving gives \(k=\frac{20\sqrt{2}}{\left(1+\frac{\pi}{4}\right)^2}\approx8.87\).

1. \([0, k]\) 2. \(t_k: y=-\frac{k\sqrt{2}}{2}x+\frac{k\sqrt{2}}{2}\left(1+\frac{\pi}{4}\right)\) 3. \(A(k)=\frac{k\sqrt{2}}{4}\left(1+\frac{\pi}{4}\right)^2\) 4. \(k=\frac{20\sqrt{2}}{\left(1+\frac{\pi}{4}\right)^2}\approx8.87\)