The function is defined for all real numbers by The minimum value of occurs at two distinct –values. What is the sum of these two –values?

Solution available with step-by-step explanation

SAT Nonlinear Functions Question 207 | Free SAT Prep | Aniko

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Question 207·Hard·Nonlinear Functions

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The function $h$ is defined for all real numbers $x \neq -1$ by

h(x) = (x + 1)^2 + \frac{9}{(x + 1)^2}.

The minimum value of $h$ occurs at two distinct $x$ –values. What is the sum of these two $x$ –values?

Your answer

(Express the answer as an integer)

For functions built from $(x+h)^2$ and its reciprocal, first simplify by substituting $t = (x+h)^2$ so the problem becomes minimizing an expression of the form $t + \frac{k}{t}$ with $t>0$ . Then use an algebraic inequality by rewriting it as a square plus a constant (for example, $(\text{something})^2 + \text{constant}$ ) to spot the minimum without calculus. Finally, translate back from $t$ to $x$ ; when the solutions are of the form $h \pm \text{something}$ , use the fact that their sum is $2h$ to get the answer quickly.

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Look for a helpful substitution

Notice that $x$ appears only as $(x+1)^2$ and in the denominator as $(x+1)^2$ . Try defining a new variable to stand for $(x+1)^2$ to simplify the expression.

Focus on minimizing a simpler expression

After you let $t = (x+1)^2$ , you get $h(x) = t + \frac{9}{t}$ with $t>0$ . Think about how to find the smallest possible value of $t + \frac{9}{t}$ using the idea that a square of a real number is always at least 0.

Convert back to x and then add

Once you know the value of $(x+1)^2$ where the minimum occurs, solve the equation $(x+1)^2 = \text{that value}$ to find the two corresponding $x$ -values, then add those two numbers together.

Desmos Guide

Graph the function

In an expression line, type h(x) = (x+1)^2 + 9/(x+1)^2 to graph the function on Desmos.

Locate the minimum points

Pan and zoom until you clearly see the lowest part of the graph. You should see two points where the graph reaches the same smallest y-value, one on each side of a vertical line.

Find and add the x-values

Click each of the two lowest points to display their coordinates and note their x-values. In a new expression line, type the sum of these two x-values (or add them by hand); that sum is what the question is asking for.

Step-by-step Explanation

Rewrite the function with a substitution

Notice that $x$ only appears inside $(x+1)^2$ and in the denominator as $(x+1)^2$ .

Let

t = (x+1)^2.

Since $(x+1)^2 \ge 0$ and $x \ne -1$ , we actually have $t > 0$ .

In terms of $t$ , the function becomes

h(x) = t + \frac{9}{t}.

Now the problem is to find the minimum value of this expression for $t > 0$ , and then relate that back to $x$ .

Minimize the expression in terms of t

We want the minimum of

t + \frac{9}{t} \quad \text{for } t>0.

Use the idea that any square is at least 0. Let $u = \sqrt{t}$ , so $u>0$ and $t = u^2$ . Then

t + \frac{9}{t} = u^2 + \frac{9}{u^2}.

Rewrite this as

u^2 + \frac{9}{u^2} = \left(u - \frac{3}{u}\right)^2 + 6.

Because a square is never negative, we have $\left(u - \frac{3}{u}\right)^2 \ge 0$ , so the smallest possible value of the expression occurs when this square is 0. That happens when

u - \frac{3}{u} = 0 \quad \Rightarrow \quad u^2 = 3.

Since $t = u^2$ , the minimum occurs when

t = 3.

Find the x-values corresponding to the minimum

We found that the minimum of $h$ occurs when

t = (x+1)^2 = 3.

Solve for $x$ by taking square roots:

x+1 = \pm \sqrt{3}.

So the two $x$ -values where $h$ reaches its minimum are

x_1 = -1 + \sqrt{3}, \quad x_2 = -1 - \sqrt{3}.

Compute the sum of the two x-values

The question asks for the sum of these two $x$ -values. Add them:

x_1 + x_2 = (-1 + \sqrt{3}) + (-1 - \sqrt{3}).

The $\sqrt{3}$ terms cancel, leaving

x_1 + x_2 = -1 - 1 = -2.

So the sum of the two $x$ -values where $h$ attains its minimum is $-2$ .

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Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation