Calculus Examples

$g (t) = \frac{t^{2}}{t^{2} + 3}$ , $[- 1, 1]$

Step 1

Find the critical points.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Find the first derivative.

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Step 1.1.1.1

Differentiate using the Quotient Rule which states that $\frac{d}{d t} [\frac{f (t)}{g (t)}]$ is $\frac{g (t) \frac{d}{d t} [f (t)] - f (t) \frac{d}{d t} [g (t)]}{{g (t)}^{2}}$ where $f (t) = t^{2}$ and $g (t) = t^{2} + 3$ .

$\frac{(t^{2} + 3) \frac{d}{d t} [t^{2}] - t^{2} \frac{d}{d t} [t^{2} + 3]}{{(t^{2} + 3)}^{2}}$

Step 1.1.1.2

Differentiate.

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Step 1.1.1.2.1

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 2$ .

$\frac{(t^{2} + 3) (2 t) - t^{2} \frac{d}{d t} [t^{2} + 3]}{{(t^{2} + 3)}^{2}}$

Step 1.1.1.2.2

Move $2$ to the left of $t^{2} + 3$ .

$\frac{2 \cdot (t^{2} + 3) t - t^{2} \frac{d}{d t} [t^{2} + 3]}{{(t^{2} + 3)}^{2}}$

Step 1.1.1.2.3

By the Sum Rule, the derivative of $t^{2} + 3$ with respect to $t$ is $\frac{d}{d t} [t^{2}] + \frac{d}{d t} [3]$ .

$\frac{2 (t^{2} + 3) t - t^{2} (\frac{d}{d t} [t^{2}] + \frac{d}{d t} [3])}{{(t^{2} + 3)}^{2}}$

Step 1.1.1.2.4

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 2$ .

$\frac{2 (t^{2} + 3) t - t^{2} (2 t + \frac{d}{d t} [3])}{{(t^{2} + 3)}^{2}}$

Step 1.1.1.2.5

Since $3$ is constant with respect to $t$ , the derivative of $3$ with respect to $t$ is $0$ .

$\frac{2 (t^{2} + 3) t - t^{2} (2 t + 0)}{{(t^{2} + 3)}^{2}}$

Step 1.1.1.2.6

Simplify the expression.

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Step 1.1.1.2.6.1

Add $2 t$ and $0$ .

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

Step 1.1.1.2.6.2

Multiply $2$ by $- 1$ .

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

Step 1.1.1.3

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

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

Step 1.1.1.4

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

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

Step 1.1.1.5

Add $1$ and $2$ .

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

Step 1.1.1.6

Simplify.

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Step 1.1.1.6.1

Apply the distributive property.

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

Step 1.1.1.6.2

Apply the distributive property.

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

Step 1.1.1.6.3

Simplify the numerator.

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Step 1.1.1.6.3.1

Simplify each term.

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Step 1.1.1.6.3.1.1

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

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Step 1.1.1.6.3.1.1.1

Move $t$ .

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

Step 1.1.1.6.3.1.1.2

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

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Step 1.1.1.6.3.1.1.2.1

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

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

Step 1.1.1.6.3.1.1.2.2

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

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

Step 1.1.1.6.3.1.1.3

Add $1$ and $2$ .

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

Step 1.1.1.6.3.1.2

Multiply $2$ by $3$ .

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

Step 1.1.1.6.3.2

Combine the opposite terms in $2 t^{3} + 6 t - 2 t^{3}$ .

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Step 1.1.1.6.3.2.1

Subtract $2 t^{3}$ from $2 t^{3}$ .

$\frac{6 t + 0}{{(t^{2} + 3)}^{2}}$

Step 1.1.1.6.3.2.2

Add $6 t$ and $0$ .

$f' (t) = \frac{6 t}{{(t^{2} + 3)}^{2}}$

Step 1.1.2

The first derivative of $g (t)$ with respect to $t$ is $\frac{6 t}{{(t^{2} + 3)}^{2}}$ .

$\frac{6 t}{{(t^{2} + 3)}^{2}}$

Step 1.2

Set the first derivative equal to $0$ then solve the equation $\frac{6 t}{{(t^{2} + 3)}^{2}} = 0$ .

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Step 1.2.1

Set the first derivative equal to $0$ .

$\frac{6 t}{{(t^{2} + 3)}^{2}} = 0$

Step 1.2.2

Set the numerator equal to zero.

$6 t = 0$

Step 1.2.3

Divide each term in $6 t = 0$ by $6$ and simplify.

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Step 1.2.3.1

Divide each term in $6 t = 0$ by $6$ .

$\frac{6 t}{6} = \frac{0}{6}$

Step 1.2.3.2

Simplify the left side.

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Step 1.2.3.2.1

Cancel the common factor of $6$ .

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Step 1.2.3.2.1.1

Cancel the common factor.

$\frac{6 t}{6} = \frac{0}{6}$

Step 1.2.3.2.1.2

Divide $t$ by $1$ .

$t = \frac{0}{6}$

Step 1.2.3.3

Simplify the right side.

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Step 1.2.3.3.1

Divide $0$ by $6$ .

$t = 0$

Step 1.3

Find the values where the derivative is undefined.

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Step 1.3.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 1.4

Evaluate $\frac{t^{2}}{t^{2} + 3}$ at each $t$ value where the derivative is $0$ or undefined.

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Step 1.4.1

Evaluate at $t = 0$ .

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Step 1.4.1.1

Substitute $0$ for $t$ .

$\frac{{(0)}^{2}}{{(0)}^{2} + 3}$

Step 1.4.1.2

Simplify.

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Step 1.4.1.2.1

Raising $0$ to any positive power yields $0$ .

$\frac{0}{0^{2} + 3}$

Step 1.4.1.2.2

Simplify the denominator.

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Step 1.4.1.2.2.1

Raising $0$ to any positive power yields $0$ .

$\frac{0}{0 + 3}$

Step 1.4.1.2.2.2

Add $0$ and $3$ .

$\frac{0}{3}$

Step 1.4.1.2.3

Divide $0$ by $3$ .

$0$

Step 1.4.2

List all of the points.

$(0, 0)$

Step 2

Evaluate at the included endpoints.

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Step 2.1

Evaluate at $t = - 1$ .

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Step 2.1.1

Substitute $- 1$ for $t$ .

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

Step 2.1.2

Simplify.

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Step 2.1.2.1

Raise $- 1$ to the power of $2$ .

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

Step 2.1.2.2

Simplify the denominator.

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Step 2.1.2.2.1

Raise $- 1$ to the power of $2$ .

$\frac{1}{1 + 3}$

Step 2.1.2.2.2

Add $1$ and $3$ .

$\frac{1}{4}$

Step 2.2

Evaluate at $t = 1$ .

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Step 2.2.1

Substitute $1$ for $t$ .

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

Step 2.2.2

Simplify.

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Step 2.2.2.1

One to any power is one.

$\frac{1}{1^{2} + 3}$

Step 2.2.2.2

Simplify the denominator.

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Step 2.2.2.2.1

One to any power is one.

$\frac{1}{1 + 3}$

Step 2.2.2.2.2

Add $1$ and $3$ .

$\frac{1}{4}$

Step 2.3

List all of the points.

$(- 1, \frac{1}{4}), (1, \frac{1}{4})$

Step 3

Compare the $g (t)$ values found for each value of $t$ in order to determine the absolute maximum and minimum over the given interval. The maximum will occur at the highest $g (t)$ value and the minimum will occur at the lowest $g (t)$ value.

Absolute Maximum: $(- 1, \frac{1}{4}), (1, \frac{1}{4})$

Absolute Minimum: $(0, 0)$

Step 4