For , the function is defined by What is the minimum value of for ?

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SAT Nonlinear Functions Question 42 | Free SAT Prep | Aniko

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Question 42·Hard·Nonlinear Functions

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For $x \ne -1$ , the function $g$ is defined by

g(x)=\frac{x^2+4x+7}{x+1}.

What is the minimum value of $g$ for $x>-1$ ?

For rational functions of the form $\dfrac{\text{quadratic}}{\text{linear}}$ on an interval, first use polynomial long division to rewrite them as a linear term plus a simpler fraction. If the denominator is $x+1$ and the domain tells you $x>-1$ , set $t=x+1>0$ to simplify. Look for expressions like $t + \dfrac{k}{t}$ , which you can minimize quickly using AM-GM or by taking a derivative. Finally, plug back to the original variable and compute the function value to choose the correct answer, rather than relying on guessing from a few test points.

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Rewrite the rational function

Try rewriting $g(x)$ by dividing $x^2+4x+7$ by $x+1$ . This often makes it easier to see how the function behaves.

Use the domain information

You are told $x>-1$ . How can you use this fact if you introduce a new variable like $t = x+1$ ?

Focus on the variable part

After substitution, you should get an expression that looks like a constant plus something of the form $t + \dfrac{4}{t}$ with $t>0$ . How can you find the smallest possible value of $t + \dfrac{4}{t}$ ?

Check where the minimum occurs

Once you find the value of $t$ that makes $t + \dfrac{4}{t}$ as small as possible, translate that back to $x$ and evaluate $g(x)$ at that $x$ .

Desmos Guide

Graph the function with its domain

In Desmos, enter the function as y = (x^2 + 4x + 7)/(x + 1). Optionally, restrict the domain by typing y = (x^2 + 4x + 7)/(x + 1) {x > -1} so the graph only shows for $x>-1$ .

Locate the minimum on the graph

Zoom in around the lowest part of the curve (you should see it somewhere to the right of $x=-1$ ). Click or tap on the curve near its lowest point so Desmos displays the coordinates. Read off the $y$ -value of this lowest point in the region $x>-1$ ; that $y$ -value is the minimum of $g$ on the given domain.

Step-by-step Explanation

Rewrite the function using division

Start by dividing the numerator by the denominator.

\frac{x^2+4x+7}{x+1}

Perform the division:

\begin{aligned} x^2+4x+7 &= (x+1)(x+3) + 4 \\ &= x^2+4x+3 + 4. \end{aligned}

g(x) = \frac{x^2+4x+7}{x+1} = x+3 + \frac{4}{x+1}.

This form separates a linear term and a simpler fraction, which is easier to analyze for minimum or maximum values.

Use a substitution to simplify the expression

Because we are told $x>-1$ , we know $x+1>0$ .

Let $t = x+1$ , so $t>0$ and $x = t-1$ .

Then

\begin{aligned} g(x) &= x+3 + \frac{4}{x+1} \\ &= (t-1)+3 + \frac{4}{t} \\ &= t+2 + \frac{4}{t}. \end{aligned}

So minimizing $g(x)$ for $x>-1$ is the same as minimizing

h(t) = t+2+\frac{4}{t} \quad \text{for } t>0.

Find the minimum of the core part $t + \frac{4}{t}$

Focus on the part of $h(t)$ that can change a lot:

F(t) = t + \frac{4}{t} \quad (t>0).

You can use the AM-GM inequality on the two positive terms $t$ and $\dfrac{4}{t}$ :

\frac{t + \frac{4}{t}}{2} \ge \sqrt{t \cdot \frac{4}{t}} = \sqrt{4} = 2.

Multiply both sides by $2$ :

t + \frac{4}{t} \ge 4.

Equality in AM-GM happens when the two terms are equal:

t = \frac{4}{t} \quad \Rightarrow \quad t^2 = 4 \quad \Rightarrow \quad t = 2,

and since $t>0$ , we take $t=2$ . So the smallest possible value of $t + \dfrac{4}{t}$ is $4$ , and it occurs when $t=2$ .

Translate back to $g(x)$ and give the minimum value

We found that for $t>0$ ,

t + \frac{4}{t} \ge 4,

h(t) = t + 2 + \frac{4}{t} \ge 4 + 2 = 6.

The minimum occurs when $t=2$ (from the previous step). Since $t=x+1$ , we have

x+1 = 2 \quad \Rightarrow \quad x = 1,

which is allowed because $x>-1$ .

At $x=1$ ,

g(1) = \frac{1^2 + 4\cdot1 + 7}{1+1} = \frac{1+4+7}{2} = \frac{12}{2} = 6.

Therefore, the minimum value of $g$ for $x>-1$ is $\boxed{6}$ .

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Desmos Guide

Step-by-step Explanation

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Desmos Guide

Step-by-step Explanation