Math

QuestionFind the value of dd if the minimum of f(x)=cosx+df(x) = \cos x + d is -6.

Studdy Solution

STEP 1

Assumptions1. The function is f(x)=cosx+df(x)=\cos x+d . The minimum value of the function is -6

STEP 2

We know that the minimum value of the cosine function cosx\cos x is -1. This occurs at x=π,π,5π,x=\pi,\pi,5\pi, \ldots and so on.
min(cosx)=1\min(\cos x) = -1

STEP 3

Since f(x)=cosx+df(x)=\cos x+d, the minimum value of f(x)f(x) is achieved when cosx\cos x is at its minimum. Therefore, the minimum value of f(x)f(x) is 1+d-1+d.
min(f(x))=min(cosx)+d=1+d\min(f(x)) = \min(\cos x) + d = -1 + d

STEP 4

We are given that the minimum value of f(x)f(x) is -6. Therefore, we can set up the equation 1+d=6-1+d=-6.
1+d=6-1 + d = -6

STEP 5

olve the equation for dd.
d=+1d = - +1

STEP 6

Calculate the value of dd.
d=6+1=5d = -6 +1 = -5The value of dd is -5.

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