Minimum value of a sin function

Question

How can I find the minimum value of this function:

\begin{equation} f(x)= \sin \left(x + \frac{π}{4} \right) + k \end{equation} I found the derivative and equated it to 0 but I got the answer wrong.

\begin{equation} 0 = \cos \left(x + \frac{π}{4} \right) \end{equation} Its a non-calculator question. How do I get to the answer using my method or another method? Thanks

Answer 1

note that $$\sin\left(x+\frac{\pi}{4}\right)$$ so you have to solve $$f'(x)=\cos\left(x+\frac{\pi}{4}\right)=0$$

Answer 2

The function $\sin(x+\pi/4)$ has minimum $-1$, therefore $\sin(x+\pi/4)+k$ has minimum $\color{red}{k-1}$.

Answer 3

Essentially, you want $\alpha=x+π/2$ so that $$\cos(\alpha)=0.$$ This occurs when $\alpha=nπ/2$, where $n$ is an odd integer.

Note that this merely gives you the points where $f(x)$ is either maximum or minimum. To find the points where the function is at its minimum you'll have to differentiate $f'(x)$ too.

For all these points of the domain where $f$ is minimum, $f=k-1.$