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Math  /  Algebra

QuestionFind the range of the function f(t)=−4cos⁡t+7f(t)=-4 \cos t+7 i ≤f(t)≤\leq f(t) \leq □\square

Studdy Solution

STEP 1

What is this asking? What are the highest and lowest values the function −4cos⁡t+7-4 \cos t + 7 can reach? Watch out! Remember that the cosine function bounces between -1 and 1!

STEP 2

1. Find the highest value.
2. Find the lowest value.

STEP 3

Alright, let's **tackle this**!
We want to find the *biggest* value that f(t)=−4cos⁡t+7f(t) = -4 \cos t + 7 can be.
Since we're subtracting 4cos⁡t4 \cos t from 7, we want cos⁡t\cos t to be as *negative* as possible to make what we're subtracting as *positive* as possible.

STEP 4

The *most negative* cos⁡t\cos t can be is **-1**.
So, let's plug that in!

STEP 5

f(t)=−4⋅(−1)+7f(t) = -4 \cdot (-1) + 7 f(t)=4+7f(t) = 4 + 7 f(t)=11f(t) = 11So, the **highest value** f(t)f(t) can reach is **11**!

STEP 6

Now, let's find the *smallest* value f(t)f(t) can be.
This time, we want cos⁡t\cos t to be as *positive* as possible to make what we're subtracting as *negative* as possible.

STEP 7

The *most positive* cos⁡t\cos t can be is **1**.
Let's plug *that* in!

STEP 8

f(t)=−4⋅(1)+7f(t) = -4 \cdot (1) + 7 f(t)=−4+7f(t) = -4 + 7 f(t)=3f(t) = 3So the **lowest value** f(t)f(t) can reach is **3**!

STEP 9

The range of f(t)f(t) is 3≤f(t)≤113 \leq f(t) \leq 11.

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