consider the given functions.\n$f(x)=31cos(x+\frac{pi}{2})+24$\n$k(x)=31cos(x-\frac{3pi}{4})+24$\nwhat is…

Explanation:

Step1: Recall the horizontal shift formula

For a function (y = A\cos(B(x - h))+C), the horizontal shift is (h) units. If (h>0), shift right; if (h < 0), shift left. The original function (f(x)=31\cos(x+\frac{\pi}{2}) + 24=31\cos(x-(-\frac{\pi}{2}))+24), and the new function (k(x)=31\cos(x-\frac{3\pi}{4})+24)

Step2: Calculate the shift amount

We want to find (h) such that (x-(-\frac{\pi}{2})+h=x - \frac{3\pi}{4}). Solving for (h): [h=-\frac{3\pi}{4}-\frac{\pi}{2}=-\frac{3\pi}{4}-\frac{2\pi}{4}=-\frac{5\pi}{4}] Another way: Let (u=x+\frac{\pi}{2}) for (f(x)) and (v=x-\frac{3\pi}{4}) for (k(x)). We want to find the change in (x) such that (v = u+h). So (x-\frac{3\pi}{4}=(x+\frac{\pi}{2})+h), then (h=-\frac{3\pi}{4}-\frac{\pi}{2}=-\frac{5\pi}{4}) (this is wrong, correct formula: if (y = f(x)) and (y = f(x - h)), for (f(x)=31\cos(x+\frac{\pi}{2})+24) and (k(x)=31\cos(x-\frac{3\pi}{4})+24), we set (x+\frac{\pi}{2}=x - \frac{3\pi}{4}-h) (no, correct: for (y = A\cos(B(x - h_1))+C) and (y = A\cos(B(x - h_2))+C), the shift is (h_2 - h_1). Here (h_1=-\frac{\pi}{2}), (h_2=\frac{3\pi}{4}), so shift (h = h_2 - h_1=\frac{3\pi}{4}-(-\frac{\pi}{2})=\frac{3\pi + 2\pi}{4}=\frac{5\pi}{4}) to the right.

Answer:

a shift right (\frac{5\pi}{4}) units