Chapter 2.4, Problem 18E

To determine

To check: Whether the curve $f\left(x\right)=\{\begin{array}{l}\sin x,0\le x<\frac{3\pi }{4}\\ \cos x,\frac{3\pi }{4}\le x\le 2\pi \end{array}$ has a tangent at $x=\frac{3\pi }{4}$ or not. If it exists give the slope of tangent.

Answer to Problem 18E

No, the curve $f\left(x\right)=\{\begin{array}{l}\sin x,0\le x<\frac{3\pi }{4}\\ \cos x,\frac{3\pi }{4}\le x\le 2\pi \end{array}$ does not have a tangent at $x=\frac{3\pi }{4}$ as the slope on the left side does not exist.

Explanation of Solution

Given information:

The curve is $f\left(x\right)=\{\begin{array}{l}\sin x,0\le x<\frac{3\pi }{4}\\ \cos x,\frac{3\pi }{4}\le x\le 2\pi \end{array}$ and the point is $x=\frac{3\pi }{4}$.

Calculation:

For the tangent of the curve $f\left(x\right)$ to exist at $x=a$ , the slope from the left side of the point $x=a$ must be equal to the slope from the right side of the point $x=a$.

The function corresponding to the left side of the point $x=\frac{3\pi }{4}$ is $f\left(x\right)=\sin x$.

The slope of the curve from the left side of $x=\frac{3\pi }{4}$ is:

  $\begin{array}{c}\underset{h\to 0^{-}}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}=\underset{h\to 0^{-}}{\lim }\frac{f\left(\frac{3\pi }{4}+h\right)-f\left(\frac{3\pi }{4}\right)}{h}\\ =\underset{h\to 0^{-}}{\lim }\frac{f\left(\frac{3\pi }{4}+h\right)-f\left(\frac{3\pi }{4}\right)}{h}\end{array}$

Substitute $\frac{3\pi }{4}+h$ for $x$ in the function $f\left(x\right)=\sin x$.

  $\begin{array}{c}f\left(\frac{3\pi }{4}+h\right)=\sin \left(\frac{3\pi }{4}+h\right)\\ f\left(\frac{3\pi }{4}+h\right)=\sin \left(\frac{3\pi }{4}\right)\cos \left(h\right)+\cos \left(\frac{3\pi }{4}\right)\sin \left(h\right)\\ f\left(\frac{3\pi }{4}+h\right)=\frac{1}{\sqrt{2}}\cos \left(h\right)-\frac{1}{\sqrt{2}}\sin \left(h\right)\end{array}$

Substitute $\frac{3\pi }{4}$ for $x$ in the function $f\left(x\right)=\cos x$.

  $\begin{array}{c}f\left(\frac{3\pi }{4}\right)=\cos \left(\frac{3\pi }{4}\right)\\ f\left(\frac{3\pi }{4}\right)=-\frac{1}{\sqrt{2}}\end{array}$

Substitute $\frac{1}{\sqrt{2}}\cos \left(h\right)-\frac{1}{\sqrt{2}}\sin \left(h\right)$ for $f\left(\frac{3\pi }{4}+h\right)$ and $-\frac{1}{\sqrt{2}}$ for $f\left(\frac{3\pi }{4}\right)$ in the formula for slope.

  $\begin{array}{c}\underset{h\to 0^{-}}{\lim }\frac{f\left(\frac{3\pi }{4}+h\right)-f\left(\frac{3\pi }{4}\right)}{h}=\underset{h\to 0^{-}}{\lim }\frac{\frac{1}{\sqrt{2}}\cos \left(h\right)-\frac{1}{\sqrt{2}}\sin \left(h\right)+\frac{1}{\sqrt{2}}}{h}\\ =\underset{h\to 0^{-}}{\lim }\left(\frac{1}{\sqrt{2}}\times \frac{\cos \left(h\right)}{h}-\left(\frac{1}{\sqrt{2}}\times \frac{\sin \left(h\right)}{h}\right)+\frac{1}{\sqrt{2}}\times \frac{1}{h}\right)\\ =\frac{1}{\sqrt{2}}\left(\underset{h\to 0^{-}}{\lim }\left(\frac{\cos \left(h\right)}{h}\right)-\underset{h\to 0^{-}}{\lim }\left(\frac{\sin \left(h\right)}{h}\right)+\underset{h\to 0^{-}}{\lim }\left(\frac{1}{h}\right)\right)\end{array}$

Further simplify.

  $\begin{array}{c}\underset{h\to 0^{-}}{\lim }\frac{f\left(\frac{3\pi }{4}+h\right)-f\left(\frac{3\pi }{4}\right)}{h}=\frac{1}{\sqrt{2}}\left(\underset{h\to 0^{-}}{\lim }\left(\frac{1+\cos \left(h\right)}{h}\right)-\underset{h\to 0^{-}}{\lim }\left(\frac{\sin \left(h\right)}{h}\right)\right)\\ =\frac{1}{\sqrt{2}}\left(\infty -1\right)\end{array}$

So, the slope of the curve from the left side of the point $x=\frac{3\pi }{4}$ does not exist.

Therefore, the curve $f\left(x\right)=\{\begin{array}{l}\sin x,0\le x<\frac{3\pi }{4}\\ \cos x,\frac{3\pi }{4}\le x\le 2\pi \end{array}$ does not have a tangent at $x=\frac{3\pi }{4}$.