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What is the minimum value of the function g(x) = |x – 4| + 8 on the interval [-2, 1]?

User LGB
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1 Answer

3 votes

Answer:


y=11

Explanation:

So we have the function:


g(x)=|x-4|+8

First, note that this is an absolute value function.

The standard form of an absolute value function is:


y=|mx-b|+c

Where m is the slope and the vertex point is (b/m, c).

Let's determine our vertex first. B is 4, m is 1, and c is 8. So:


(4/1,8)=(4,8)

So, our vertex is at (4,8).

Also, note that there is no negative in front of our absolute value, so this is going upwards.

Now, recall that an absolute value function has the general shape of a V. Since this is a positive absolute value, all x-values to the left of the vertex will be decreasing, while all x-values to the right of the vertex will be increasing.

In other words, from negative infinity to 4, the function will be decreasing.

And from 4 until positive infinity, the function will be increasing.

We want to find the minimum value of the function between [-2,1].

Since [-2,1] is to the left of the vertex, this means that the function is decreasing.

So, to find the minimum value, we just need to use the x-value closest to the vertex, in this case, x=1.

So, to find the minimum value, we just need to substitute 1 into g(x). So:


g(1)=|1-4|+8

Subtract:


g(1)=|-3|+8

Absolute Value:


g(1)=3+8

Add:


g(1)=11

Therefore, our minimum value of the g(x) on the interval [-2,1] is y=11.

We can graph the function to verify our answer. As we can see, the minimum value on the interval [-2,1] (black line) is indeed y=11.

What is the minimum value of the function g(x) = |x – 4| + 8 on the interval [-2, 1]?-example-1
User Youssoua
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