Q. 51 - Calculus

Question

Use a sign chart for $f\text{'}\text{'}$ to determine the intervals on which each function f in Exercises 41–52 is concave up or concave down, and identify the locations of any inflection points. Then verify your algebraic answers with graphs from a calculator or graphing utility 

$f\left(x\right)=\sin \left(x-\frac{\pi }{4}\right)$

Step-by-Step Solution

Answer

Inflection points are at$x=\frac{\pi }{4}+2\pi k,\frac{5\pi }{4}+2\pi k$, concave up at all other places other than where it is concave down that is at $\left(\frac{\pi }{4}+2\pi k,\frac{5\pi }{4}+2\pi k\right).$

Step 1. Given information.

The given function is $f\left(x\right)=\sin \left(x-\frac{\pi }{4}\right).$

Step 2. Second derivative.

On differentiating the function, we get,

$$\begin{gathered} f^{\text{'}}\left(x\right)=\frac{d}{dx}\sin \left(x-\frac{\pi }{4}\right) \\ =\cos \left(x-\frac{\pi }{4}\right) \\ {f}^{\text{'}\text{'}}\left(x\right)=\frac{d}{dx}\cos \left(x-\frac{\pi }{4}\right) \\ =-\sin \left(x-\frac{\pi }{4}\right) \end{gathered}$$

Step 3. Sign chart.

Now,

$f\text{'}\text{'}\left(x\right)=0 at x=\frac{\pi }{4}+2\pi k,\frac{5\pi }{4}+2\pi k$

Therefore, the chart will look like,

Inflection point is $x=\frac{\pi }{4}+2\pi k,\frac{5\pi }{4}+2\pi k$ and concave up at all other places other than where it is concave down that is at $\left(\frac{\pi }{4}+2\pi k,\frac{5\pi }{4}+2\pi k\right).$

Step 4. Verification.

The graph of the function is,