Calculus Examples

$t^{4} + t^{3} + t^{2} + 1$

Step 1

Find the first derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

By the Sum Rule, the derivative of $t^{4} + t^{3} + t^{2} + 1$ with respect to $t$ is $\frac{d}{d t} [t^{4}] + \frac{d}{d t} [t^{3}] + \frac{d}{d t} [t^{2}] + \frac{d}{d t} [1]$ .

$\frac{d}{d t} [t^{4}] + \frac{d}{d t} [t^{3}] + \frac{d}{d t} [t^{2}] + \frac{d}{d t} [1]$

Step 1.1.2

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 4$ .

$4 t^{3} + \frac{d}{d t} [t^{3}] + \frac{d}{d t} [t^{2}] + \frac{d}{d t} [1]$

Step 1.1.3

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 3$ .

$4 t^{3} + 3 t^{2} + \frac{d}{d t} [t^{2}] + \frac{d}{d t} [1]$

Step 1.1.4

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

$4 t^{3} + 3 t^{2} + 2 t + \frac{d}{d t} [1]$

Step 1.1.5

Since $1$ is constant with respect to $t$ , the derivative of $1$ with respect to $t$ is $0$ .

$4 t^{3} + 3 t^{2} + 2 t + 0$

Step 1.1.6

Add $4 t^{3} + 3 t^{2} + 2 t$ and $0$ .

$f' (t) = 4 t^{3} + 3 t^{2} + 2 t$

Step 1.2

The first derivative of $f (t)$ with respect to $t$ is $4 t^{3} + 3 t^{2} + 2 t$ .

$4 t^{3} + 3 t^{2} + 2 t$

Step 2

Set the first derivative equal to $0$ then solve the equation $4 t^{3} + 3 t^{2} + 2 t = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$4 t^{3} + 3 t^{2} + 2 t = 0$

Step 2.2

Factor $t$ out of $4 t^{3} + 3 t^{2} + 2 t$ .

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Step 2.2.1

Factor $t$ out of $4 t^{3}$ .

$t (4 t^{2}) + 3 t^{2} + 2 t = 0$

Step 2.2.2

Factor $t$ out of $3 t^{2}$ .

$t (4 t^{2}) + t (3 t) + 2 t = 0$

Step 2.2.3

Factor $t$ out of $2 t$ .

$t (4 t^{2}) + t (3 t) + t \cdot 2 = 0$

Step 2.2.4

Factor $t$ out of $t (4 t^{2}) + t (3 t)$ .

$t (4 t^{2} + 3 t) + t \cdot 2 = 0$

Step 2.2.5

Factor $t$ out of $t (4 t^{2} + 3 t) + t \cdot 2$ .

$t (4 t^{2} + 3 t + 2) = 0$

Step 2.3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$t = 0$

$4 t^{2} + 3 t + 2 = 0$

Step 2.4

Set $t$ equal to $0$ .

$t = 0$

Step 2.5

Set $4 t^{2} + 3 t + 2$ equal to $0$ and solve for $t$ .

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Step 2.5.1

Set $4 t^{2} + 3 t + 2$ equal to $0$ .

$4 t^{2} + 3 t + 2 = 0$

Step 2.5.2

Solve $4 t^{2} + 3 t + 2 = 0$ for $t$ .

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Step 2.5.2.1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2.5.2.2

Substitute the values $a = 4$ , $b = 3$ , and $c = 2$ into the quadratic formula and solve for $t$ .

$\frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (4 \cdot 2)}}{2 \cdot 4}$

Step 2.5.2.3

Simplify.

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Step 2.5.2.3.1

Simplify the numerator.

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Step 2.5.2.3.1.1

Raise $3$ to the power of $2$ .

$t = \frac{- 3 \pm \sqrt{9 - 4 \cdot 4 \cdot 2}}{2 \cdot 4}$

Step 2.5.2.3.1.2

Multiply $- 4 \cdot 4 \cdot 2$ .

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Step 2.5.2.3.1.2.1

Multiply $- 4$ by $4$ .

$t = \frac{- 3 \pm \sqrt{9 - 16 \cdot 2}}{2 \cdot 4}$

Step 2.5.2.3.1.2.2

Multiply $- 16$ by $2$ .

$t = \frac{- 3 \pm \sqrt{9 - 32}}{2 \cdot 4}$

Step 2.5.2.3.1.3

Subtract $32$ from $9$ .

$t = \frac{- 3 \pm \sqrt{- 23}}{2 \cdot 4}$

Step 2.5.2.3.1.4

Rewrite $- 23$ as $- 1 (23)$ .

$t = \frac{- 3 \pm \sqrt{- 1 \cdot 23}}{2 \cdot 4}$

Step 2.5.2.3.1.5

Rewrite $\sqrt{- 1 (23)}$ as $\sqrt{- 1} \cdot \sqrt{23}$ .

$t = \frac{- 3 \pm \sqrt{- 1} \cdot \sqrt{23}}{2 \cdot 4}$

Step 2.5.2.3.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$t = \frac{- 3 \pm i \sqrt{23}}{2 \cdot 4}$

Step 2.5.2.3.2

Multiply $2$ by $4$ .

$t = \frac{- 3 \pm i \sqrt{23}}{8}$

Step 2.5.2.4

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 2.5.2.4.1

Simplify the numerator.

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Step 2.5.2.4.1.1

Raise $3$ to the power of $2$ .

$t = \frac{- 3 \pm \sqrt{9 - 4 \cdot 4 \cdot 2}}{2 \cdot 4}$

Step 2.5.2.4.1.2

Multiply $- 4 \cdot 4 \cdot 2$ .

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Step 2.5.2.4.1.2.1

Multiply $- 4$ by $4$ .

$t = \frac{- 3 \pm \sqrt{9 - 16 \cdot 2}}{2 \cdot 4}$

Step 2.5.2.4.1.2.2

Multiply $- 16$ by $2$ .

$t = \frac{- 3 \pm \sqrt{9 - 32}}{2 \cdot 4}$

Step 2.5.2.4.1.3

Subtract $32$ from $9$ .

$t = \frac{- 3 \pm \sqrt{- 23}}{2 \cdot 4}$

Step 2.5.2.4.1.4

Rewrite $- 23$ as $- 1 (23)$ .

$t = \frac{- 3 \pm \sqrt{- 1 \cdot 23}}{2 \cdot 4}$

Step 2.5.2.4.1.5

Rewrite $\sqrt{- 1 (23)}$ as $\sqrt{- 1} \cdot \sqrt{23}$ .

$t = \frac{- 3 \pm \sqrt{- 1} \cdot \sqrt{23}}{2 \cdot 4}$

Step 2.5.2.4.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$t = \frac{- 3 \pm i \sqrt{23}}{2 \cdot 4}$

Step 2.5.2.4.2

Multiply $2$ by $4$ .

$t = \frac{- 3 \pm i \sqrt{23}}{8}$

Step 2.5.2.4.3

Change the $\pm$ to $+$ .

$t = \frac{- 3 + i \sqrt{23}}{8}$

Step 2.5.2.4.4

Rewrite $- 3$ as $- 1 (3)$ .

$t = \frac{- 1 \cdot 3 + i \sqrt{23}}{8}$

Step 2.5.2.4.5

Factor $- 1$ out of $i \sqrt{23}$ .

$t = \frac{- 1 \cdot 3 - (- i \sqrt{23})}{8}$

Step 2.5.2.4.6

Factor $- 1$ out of $- 1 (3) - (- i \sqrt{23})$ .

$t = \frac{- 1 (3 - i \sqrt{23})}{8}$

Step 2.5.2.4.7

Move the negative in front of the fraction.

$t = - \frac{3 - i \sqrt{23}}{8}$

Step 2.5.2.5

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 2.5.2.5.1

Simplify the numerator.

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Step 2.5.2.5.1.1

Raise $3$ to the power of $2$ .

$t = \frac{- 3 \pm \sqrt{9 - 4 \cdot 4 \cdot 2}}{2 \cdot 4}$

Step 2.5.2.5.1.2

Multiply $- 4 \cdot 4 \cdot 2$ .

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Step 2.5.2.5.1.2.1

Multiply $- 4$ by $4$ .

$t = \frac{- 3 \pm \sqrt{9 - 16 \cdot 2}}{2 \cdot 4}$

Step 2.5.2.5.1.2.2

Multiply $- 16$ by $2$ .

$t = \frac{- 3 \pm \sqrt{9 - 32}}{2 \cdot 4}$

Step 2.5.2.5.1.3

Subtract $32$ from $9$ .

$t = \frac{- 3 \pm \sqrt{- 23}}{2 \cdot 4}$

Step 2.5.2.5.1.4

Rewrite $- 23$ as $- 1 (23)$ .

$t = \frac{- 3 \pm \sqrt{- 1 \cdot 23}}{2 \cdot 4}$

Step 2.5.2.5.1.5

Rewrite $\sqrt{- 1 (23)}$ as $\sqrt{- 1} \cdot \sqrt{23}$ .

$t = \frac{- 3 \pm \sqrt{- 1} \cdot \sqrt{23}}{2 \cdot 4}$

Step 2.5.2.5.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$t = \frac{- 3 \pm i \sqrt{23}}{2 \cdot 4}$

Step 2.5.2.5.2

Multiply $2$ by $4$ .

$t = \frac{- 3 \pm i \sqrt{23}}{8}$

Step 2.5.2.5.3

Change the $\pm$ to $-$ .

$t = \frac{- 3 - i \sqrt{23}}{8}$

Step 2.5.2.5.4

Rewrite $- 3$ as $- 1 (3)$ .

$t = \frac{- 1 \cdot 3 - i \sqrt{23}}{8}$

Step 2.5.2.5.5

Factor $- 1$ out of $- i \sqrt{23}$ .

$t = \frac{- 1 \cdot 3 - (i \sqrt{23})}{8}$

Step 2.5.2.5.6

Factor $- 1$ out of $- 1 (3) - (i \sqrt{23})$ .

$t = \frac{- 1 (3 + i \sqrt{23})}{8}$

Step 2.5.2.5.7

Move the negative in front of the fraction.

$t = - \frac{3 + i \sqrt{23}}{8}$

Step 2.5.2.6

The final answer is the combination of both solutions.

$t = - \frac{3 - i \sqrt{23}}{8}, - \frac{3 + i \sqrt{23}}{8}$

Step 2.6

The final solution is all the values that make $t (4 t^{2} + 3 t + 2) = 0$ true.

$t = 0, - \frac{3 - i \sqrt{23}}{8}, - \frac{3 + i \sqrt{23}}{8}$

Step 3

Find the values where the derivative is undefined.

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Step 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 4

Evaluate $t^{4} + t^{3} + t^{2} + 1$ at each $t$ value where the derivative is $0$ or undefined.

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Step 4.1

Evaluate at $t = 0$ .

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Step 4.1.1

Substitute $0$ for $t$ .

${(0)}^{4} + {(0)}^{3} + {(0)}^{2} + 1$

Step 4.1.2

Simplify.

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Step 4.1.2.1

Simplify each term.

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Step 4.1.2.1.1

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

$0 + {(0)}^{3} + {(0)}^{2} + 1$

Step 4.1.2.1.2

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

$0 + 0 + {(0)}^{2} + 1$

Step 4.1.2.1.3

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

$0 + 0 + 0 + 1$

Step 4.1.2.2

Simplify by adding numbers.

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Step 4.1.2.2.1

Add $0$ and $0$ .

$0 + 0 + 1$

Step 4.1.2.2.2

Add $0$ and $0$ .

$0 + 1$

Step 4.1.2.2.3

Add $0$ and $1$ .

$1$

Step 4.2

List all of the points.

$(0, 1)$

Step 5