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Problem 168

# If $$\mathrm{g}(\mathrm{x})=\mathrm{x}^{2}-2 \mathrm{x}+1$$, find the given element $$\mathrm{m}$$ the range, a) $$g(-2)$$ b) $$\mathrm{g}(0)$$ c) $$g(a+1)$$ d) $$g(a-1)$$

Expert verified
The values of g(x) for the given input values are: a) $$g(-2) = 9$$ b) $$g(0) = 1$$ c) $$g(a+1) = a^2$$ d) $$g(a-1) = a^2 - 4a + 4$$
See the step by step solution

## Step 1: Find g(-2)

First, let's find the value of g(-2) by substituting -2 in place of x in the given formula: g(-2) = (-2)^2 - 2(-2) + 1 g(-2) = 4 + 4 + 1 g(-2) = 9 So, g(-2) = 9.

## Step 2: Find g(0)

Next, let's find the value of g(0) by substituting 0 in place of x: g(0) = (0)^2 - 2(0) + 1 g(0) = 0 + 0 + 1 g(0) = 1 So, g(0) = 1.

## Step 3: Find g(a+1)

Now, let's find the value of g(a+1) by substituting a+1 in place of x: g(a+1) = (a+1)^2 - 2(a+1) + 1 g(a+1) = (a^2 + 2a + 1) - 2(a + 1) + 1 g(a+1) = a^2 + 2a + 1 - 2a - 2 + 1 g(a+1) = a^2 So, g(a+1) = a^2.

## Step 4: Find g(a-1)

Lastly, let's find the value of g(a-1) by substituting a-1 in place of x: g(a-1) = (a-1)^2 - 2(a-1) + 1 g(a-1) = (a^2 - 2a + 1) - 2(a - 1) + 1 g(a-1) = a^2 - 2a + 1 - 2a + 2 + 1 g(a-1) = a^2 - 4a + 4 So, g(a-1) = a^2 - 4a + 4. Now, we have found the values of g(x) for all given input values: a) g(-2) = 9 b) g(0) = 1 c) g(a+1) = a^2 d) g(a-1) = a^2 - 4a + 4

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