Find the Domain of the Difference of the Two Functions f(t)=t^2+7t+1 ; g(t)=2/(t-1)

$f (t) = t^{2} + 7 t + 1$ ; $g (t) = \frac{2}{t - 1}$

Step 1

Find the difference of the functions.

Replace the function designators with the actual functions in $f (t) - g t$ .

$t^{2} + 7 t + 1 - (\frac{2}{t - 1})$

To write $t^{2}$ as a fraction with a common denominator, multiply by $\frac{t - 1}{t - 1}$ .

$7 t + 1 + \frac{t^{2} (t - 1)}{t - 1} - \frac{2}{t - 1}$

Combine the numerators over the common denominator.

$7 t + 1 + \frac{t^{2} (t - 1) - 2}{t - 1}$

Simplify the numerator.

Apply the distributive property.

$7 t + 1 + \frac{t^{2} t + t^{2} \cdot - 1 - 2}{t - 1}$

Multiply $t^{2}$ by $t$ by adding the exponents.

Multiply $t^{2}$ by $t$ .

Raise $t$ to the power of $1$ .

$7 t + 1 + \frac{t^{2} t^{1} + t^{2} \cdot - 1 - 2}{t - 1}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$7 t + 1 + \frac{t^{2 + 1} + t^{2} \cdot - 1 - 2}{t - 1}$

Add $2$ and $1$ .

$7 t + 1 + \frac{t^{3} + t^{2} \cdot - 1 - 2}{t - 1}$

Move $- 1$ to the left of $t^{2}$ .

$7 t + 1 + \frac{t^{3} - 1 \cdot t^{2} - 2}{t - 1}$

Rewrite $- 1 t^{2}$ as $- t^{2}$ .

$7 t + 1 + \frac{t^{3} - t^{2} - 2}{t - 1}$

To write $7 t$ as a fraction with a common denominator, multiply by $\frac{t - 1}{t - 1}$ .

$\frac{7 t (t - 1)}{t - 1} + \frac{t^{3} - t^{2} - 2}{t - 1} + 1$

Simplify terms.

Combine the numerators over the common denominator.

$\frac{7 t (t - 1) + t^{3} - t^{2} - 2}{t - 1} + 1$

Simplify the numerator.

Apply the distributive property.

$\frac{7 t \cdot t + 7 t \cdot - 1 + t^{3} - t^{2} - 2}{t - 1} + 1$

Multiply $t$ by $t$ by adding the exponents.

Move $t$ .

$\frac{7 (t \cdot t) + 7 t \cdot - 1 + t^{3} - t^{2} - 2}{t - 1} + 1$

Multiply $t$ by $t$ .

$\frac{7 t^{2} + 7 t \cdot - 1 + t^{3} - t^{2} - 2}{t - 1} + 1$

Multiply $- 1$ by $7$ .

$\frac{7 t^{2} - 7 t + t^{3} - t^{2} - 2}{t - 1} + 1$

Subtract $t^{2}$ from $7 t^{2}$ .

$\frac{6 t^{2} - 7 t + t^{3} - 2}{t - 1} + 1$

Reorder terms.

$\frac{t^{3} + 6 t^{2} - 7 t - 2}{t - 1} + 1$

Combine into one fraction.

Write $1$ as a fraction with a common denominator.

$\frac{t^{3} + 6 t^{2} - 7 t - 2}{t - 1} + \frac{t - 1}{t - 1}$

Combine the numerators over the common denominator.

$\frac{t^{3} + 6 t^{2} - 7 t - 2 + t - 1}{t - 1}$

Simplify the numerator.

Add $- 7 t$ and $t$ .

$\frac{t^{3} + 6 t^{2} - 6 t - 2 - 1}{t - 1}$

Subtract $1$ from $- 2$ .

$\frac{t^{3} + 6 t^{2} - 6 t - 3}{t - 1}$

Step 2

Set the denominator in $\frac{t^{3} + 6 t^{2} - 6 t - 3}{t - 1}$ equal to $0$ to find where the expression is undefined.

$t - 1 = 0$

Step 3

Add $1$ to both sides of the equation.

$t = 1$

Step 4

The domain is all values of $t$ that make the expression defined.

Interval Notation:

$(- \infty, 1) \cup (1, \infty)$

Set-Builder Notation:

${t | t \neq 1}$

Step 5

Find the Domain of the Difference of the Two Functions f(t)=t^2+7t+1 ; g(t)=2/(t-1)

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