Calculus Examples

$g (t) = (4 - t) \cdot 4^{t}$

Step 1

Find the first derivative of the function.

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Step 1.1

Differentiate using the Product Rule which states that $\frac{d}{d t} [f (t) g (t)]$ is $f (t) \frac{d}{d t} [g (t)] + g (t) \frac{d}{d t} [f (t)]$ where $f (t) = 4 - t$ and $g (t) = 4^{t}$ .

$(4 - t) \frac{d}{d t} [4^{t}] + 4^{t} \frac{d}{d t} [4 - t]$

Step 1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d t} [a^{t}]$ is $a^{t} \ln (a)$ where $a$ = $4$ .

$(4 - t) (4^{t} \ln (4)) + 4^{t} \frac{d}{d t} [4 - t]$

Step 1.3

Differentiate.

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Step 1.3.1

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

$(4 - t) \cdot 4^{t} \ln (4) + 4^{t} (\frac{d}{d t} [4] + \frac{d}{d t} [- t])$

Step 1.3.2

Since $4$ is constant with respect to $t$ , the derivative of $4$ with respect to $t$ is $0$ .

$(4 - t) \cdot 4^{t} \ln (4) + 4^{t} (0 + \frac{d}{d t} [- t])$

Step 1.3.3

Add $0$ and $\frac{d}{d t} [- t]$ .

$(4 - t) \cdot 4^{t} \ln (4) + 4^{t} \frac{d}{d t} [- t]$

Step 1.3.4

Since $- 1$ is constant with respect to $t$ , the derivative of $- t$ with respect to $t$ is $- \frac{d}{d t} [t]$ .

$(4 - t) \cdot 4^{t} \ln (4) + 4^{t} (- \frac{d}{d t} [t])$

Step 1.3.5

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

$(4 - t) \cdot 4^{t} \ln (4) + 4^{t} (- 1 \cdot 1)$

Step 1.3.6

Simplify the expression.

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Step 1.3.6.1

Multiply $- 1$ by $1$ .

$(4 - t) \cdot 4^{t} \ln (4) + 4^{t} \cdot - 1$

Step 1.3.6.2

Move $- 1$ to the left of $4^{t}$ .

$(4 - t) \cdot 4^{t} \ln (4) - 1 \cdot 4^{t}$

Step 1.3.6.3

Rewrite $- 1 \cdot 4^{t}$ as $- 4^{t}$ .

$(4 - t) \cdot 4^{t} \ln (4) - 4^{t}$

Step 1.4

Simplify.

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Step 1.4.1

Apply the distributive property.

$(4 \cdot 4^{t} - t \cdot 4^{t}) \ln (4) - 4^{t}$

Step 1.4.2

Apply the distributive property.

$4 \cdot 4^{t} \ln (4) - t \cdot 4^{t} \ln (4) - 4^{t}$

Step 1.4.3

Multiply $4$ by $4^{t}$ .

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Step 1.4.3.1

Raise $4$ to the power of $1$ .

$4^{1} \cdot 4^{t} \ln (4) - t \cdot 4^{t} \ln (4) - 4^{t}$

Step 1.4.3.2

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

$4^{1 + t} \ln (4) - t \cdot 4^{t} \ln (4) - 4^{t}$

Step 1.4.4

Reorder terms.

$4^{1 + t} \ln (4) - 4^{t} t \ln (4) - 4^{t}$

Step 2

Find the second derivative of the function.

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Step 2.1

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

$g'' (t) = \frac{d}{d t} (4^{1 + t} \ln (4)) + \frac{d}{d t} (- 4^{t} t \ln (4)) + \frac{d}{d t} (- 4^{t})$

Step 2.2

Evaluate $\frac{d}{d t} [4^{1 + t} \ln (4)]$ .

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Step 2.2.1

Since $\ln (4)$ is constant with respect to $t$ , the derivative of $4^{1 + t} \ln (4)$ with respect to $t$ is $\ln (4) \frac{d}{d t} [4^{1 + t}]$ .

$g'' (t) = \ln (4) \frac{d}{d t} (4^{1 + t}) + \frac{d}{d t} (- 4^{t} t \ln (4)) + \frac{d}{d t} (- 4^{t})$

Step 2.2.2

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = 4^{t}$ and $g (t) = 1 + t$ .

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Step 2.2.2.1

To apply the Chain Rule, set $u$ as $1 + t$ .

$g'' (t) = \ln (4) (\frac{d}{d u} (4^{u}) \frac{d}{d t} (1 + t)) + \frac{d}{d t} (- 4^{t} t \ln (4)) + \frac{d}{d t} (- 4^{t})$

Step 2.2.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $4$ .

$g'' (t) = \ln (4) (4^{u} \ln (4) \frac{d}{d t} (1 + t)) + \frac{d}{d t} (- 4^{t} t \ln (4)) + \frac{d}{d t} (- 4^{t})$

Step 2.2.2.3

Replace all occurrences of $u$ with $1 + t$ .

$g'' (t) = \ln (4) (4^{1 + t} \ln (4) \frac{d}{d t} (1 + t)) + \frac{d}{d t} (- 4^{t} t \ln (4)) + \frac{d}{d t} (- 4^{t})$

Step 2.2.3

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

$g'' (t) = \ln (4) (4^{1 + t} \ln (4) (\frac{d}{d t} (1) + \frac{d}{d t} (t))) + \frac{d}{d t} (- 4^{t} t \ln (4)) + \frac{d}{d t} (- 4^{t})$

Step 2.2.4

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

$g'' (t) = \ln (4) (4^{1 + t} \ln (4) (0 + \frac{d}{d t} (t))) + \frac{d}{d t} (- 4^{t} t \ln (4)) + \frac{d}{d t} (- 4^{t})$

Step 2.2.5

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

$g'' (t) = \ln (4) (4^{1 + t} \ln (4) (0 + 1)) + \frac{d}{d t} (- 4^{t} t \ln (4)) + \frac{d}{d t} (- 4^{t})$

Step 2.2.6

Add $0$ and $1$ .

$g'' (t) = \ln (4) (4^{1 + t} \ln (4) \cdot 1) + \frac{d}{d t} (- 4^{t} t \ln (4)) + \frac{d}{d t} (- 4^{t})$

Step 2.2.7

Multiply $4^{1 + t}$ by $1$ .

$g'' (t) = \ln (4) (4^{1 + t} \ln (4)) + \frac{d}{d t} (- 4^{t} t \ln (4)) + \frac{d}{d t} (- 4^{t})$

Step 2.2.8

Raise $\ln (4)$ to the power of $1$ .

$g'' (t) = 4^{1 + t} ((\ln (4)) \ln (4)) + \frac{d}{d t} (- 4^{t} t \ln (4)) + \frac{d}{d t} (- 4^{t})$

Step 2.2.9

Raise $\ln (4)$ to the power of $1$ .

$g'' (t) = 4^{1 + t} ((\ln (4)) (\ln (4))) + \frac{d}{d t} (- 4^{t} t \ln (4)) + \frac{d}{d t} (- 4^{t})$

Step 2.2.10

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

$g'' (t) = 4^{1 + t} {(\ln (4))}^{1 + 1} + \frac{d}{d t} (- 4^{t} t \ln (4)) + \frac{d}{d t} (- 4^{t})$

Step 2.2.11

Add $1$ and $1$ .

$g'' (t) = 4^{1 + t} \ln^{2} (4) + \frac{d}{d t} (- 4^{t} t \ln (4)) + \frac{d}{d t} (- 4^{t})$

Step 2.3

Evaluate $\frac{d}{d t} [- 4^{t} t \ln (4)]$ .

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Step 2.3.1

Since $- \ln (4)$ is constant with respect to $t$ , the derivative of $- 4^{t} t \ln (4)$ with respect to $t$ is $- \ln (4) \frac{d}{d t} [4^{t} t]$ .

$g'' (t) = 4^{1 + t} \ln^{2} (4) - \ln (4) \frac{d}{d t} (4^{t} t) + \frac{d}{d t} (- 4^{t})$

Step 2.3.2

Differentiate using the Product Rule which states that $\frac{d}{d t} [f (t) g (t)]$ is $f (t) \frac{d}{d t} [g (t)] + g (t) \frac{d}{d t} [f (t)]$ where $f (t) = 4^{t}$ and $g (t) = t$ .

$g'' (t) = 4^{1 + t} \ln^{2} (4) - \ln (4) (4^{t} \frac{d}{d t} (t) + t \frac{d}{d t} (4^{t})) + \frac{d}{d t} (- 4^{t})$

Step 2.3.3

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

$g'' (t) = 4^{1 + t} \ln^{2} (4) - \ln (4) (4^{t} \cdot 1 + t \frac{d}{d t} (4^{t})) + \frac{d}{d t} (- 4^{t})$

Step 2.3.4

Differentiate using the Exponential Rule which states that $\frac{d}{d t} [a^{t}]$ is $a^{t} \ln (a)$ where $a$ = $4$ .

$g'' (t) = 4^{1 + t} \ln^{2} (4) - \ln (4) (4^{t} \cdot 1 + t (4^{t} \ln (4))) + \frac{d}{d t} (- 4^{t})$

Step 2.3.5

Multiply $4^{t}$ by $1$ .

$g'' (t) = 4^{1 + t} \ln^{2} (4) - \ln (4) (4^{t} + t (4^{t} \ln (4))) + \frac{d}{d t} (- 4^{t})$

Step 2.4

Evaluate $\frac{d}{d t} [- 4^{t}]$ .

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Step 2.4.1

Since $- 1$ is constant with respect to $t$ , the derivative of $- 4^{t}$ with respect to $t$ is $- \frac{d}{d t} [4^{t}]$ .

$g'' (t) = 4^{1 + t} \ln^{2} (4) - \ln (4) (4^{t} + t (4^{t} \ln (4))) - \frac{d}{d t} \cdot 4^{t}$

Step 2.4.2

Differentiate using the Exponential Rule which states that $\frac{d}{d t} [a^{t}]$ is $a^{t} \ln (a)$ where $a$ = $4$ .

$g'' (t) = 4^{1 + t} \ln^{2} (4) - \ln (4) (4^{t} + t (4^{t} \ln (4))) - 4^{t} \ln (4)$

Step 2.5

Simplify.

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Step 2.5.1

Apply the distributive property.

$g'' (t) = 4^{1 + t} \ln^{2} (4) - \ln (4) \cdot 4^{t} - \ln (4) (t (4^{t} \ln (4))) - 4^{t} \ln (4)$

Step 2.5.2

Combine terms.

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Step 2.5.2.1

Raise $\ln (4)$ to the power of $1$ .

$g'' (t) = 4^{1 + t} \ln^{2} (4) - \ln (4) \cdot 4^{t} - (t (4^{t} ((\ln (4)) \ln (4)))) - 4^{t} \ln (4)$

Step 2.5.2.2

Raise $\ln (4)$ to the power of $1$ .

$g'' (t) = 4^{1 + t} \ln^{2} (4) - \ln (4) \cdot 4^{t} - (t (4^{t} ((\ln (4)) (\ln (4))))) - 4^{t} \ln (4)$

Step 2.5.2.3

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

$g'' (t) = 4^{1 + t} \ln^{2} (4) - \ln (4) \cdot 4^{t} - (t (4^{t} {(\ln (4))}^{1 + 1})) - 4^{t} \ln (4)$

Step 2.5.2.4

Add $1$ and $1$ .

$g'' (t) = 4^{1 + t} \ln^{2} (4) - \ln (4) \cdot 4^{t} - (t (4^{t} \ln^{2} (4))) - 4^{t} \ln (4)$

Step 2.5.2.5

Subtract $4^{t} \ln (4)$ from $- \ln (4) \cdot 4^{t}$ .

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Step 2.5.2.5.1

Move $\ln (4)$ .

$g'' (t) = 4^{1 + t} \ln^{2} (4) - 1 \cdot (4^{t} \ln (4)) - 4^{t} \ln (4) - t (4^{t} \ln^{2} (4))$

Step 2.5.2.5.2

Subtract $4^{t} \ln (4)$ from $- 1 \cdot 4^{t} \ln (4)$ .

$g'' (t) = 4^{1 + t} \ln^{2} (4) - 2 \cdot (4^{t} \ln (4)) - t (4^{t} \ln^{2} (4))$

Step 2.5.2.6

Factor out negative.

$g'' (t) = 4^{1 + t} \ln^{2} (4) - 2 \cdot 4^{t} \ln (4) - t (4^{t} \ln^{2} (4))$

Step 2.5.2.7

Rewrite $4$ as $2^{2}$ .

$g'' (t) = 4^{1 + t} \ln^{2} (4) - 2 \cdot {(2^{2})}^{t} \ln (4) - t (4^{t} \ln^{2} (4))$

Step 2.5.2.8

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$g'' (t) = 4^{1 + t} \ln^{2} (4) - 2 \cdot 2^{2 t} \ln (4) - t (4^{t} \ln^{2} (4))$

Step 2.5.2.9

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

$g'' (t) = 4^{1 + t} \ln^{2} (4) - 2^{1 + 2 t} \ln (4) - t (4^{t} \ln^{2} (4))$

Step 2.5.3

Reorder terms.

$g'' (t) = 4^{1 + t} \ln^{2} (4) - 2^{1 + 2 t} \ln (4) - 4^{t} t \ln^{2} (4)$

Step 2.5.4

Reorder factors in $4^{1 + t} \ln^{2} (4) - 2^{1 + 2 t} \ln (4) - 4^{t} t \ln^{2} (4)$ .

$g'' (t) = 4^{1 + t} \ln^{2} (4) - 2^{1 + 2 t} \ln (4) - t \cdot (4^{t} \ln^{2} (4))$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$4^{1 + t} \ln (4) - 4^{t} t \ln (4) - 4^{t} = 0$

Step 4

Find the first derivative.

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Step 4.1

Find the first derivative.

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Step 4.1.1

Differentiate using the Product Rule which states that $\frac{d}{d t} [f (t) g (t)]$ is $f (t) \frac{d}{d t} [g (t)] + g (t) \frac{d}{d t} [f (t)]$ where $f (t) = 4 - t$ and $g (t) = 4^{t}$ .

$(4 - t) \frac{d}{d t} [4^{t}] + 4^{t} \frac{d}{d t} [4 - t]$

Step 4.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d t} [a^{t}]$ is $a^{t} \ln (a)$ where $a$ = $4$ .

$(4 - t) (4^{t} \ln (4)) + 4^{t} \frac{d}{d t} [4 - t]$

Step 4.1.3

Differentiate.

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Step 4.1.3.1

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

$(4 - t) \cdot 4^{t} \ln (4) + 4^{t} (\frac{d}{d t} [4] + \frac{d}{d t} [- t])$

Step 4.1.3.2

Since $4$ is constant with respect to $t$ , the derivative of $4$ with respect to $t$ is $0$ .

$(4 - t) \cdot 4^{t} \ln (4) + 4^{t} (0 + \frac{d}{d t} [- t])$

Step 4.1.3.3

Add $0$ and $\frac{d}{d t} [- t]$ .

$(4 - t) \cdot 4^{t} \ln (4) + 4^{t} \frac{d}{d t} [- t]$

Step 4.1.3.4

Since $- 1$ is constant with respect to $t$ , the derivative of $- t$ with respect to $t$ is $- \frac{d}{d t} [t]$ .

$(4 - t) \cdot 4^{t} \ln (4) + 4^{t} (- \frac{d}{d t} [t])$

Step 4.1.3.5

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

$(4 - t) \cdot 4^{t} \ln (4) + 4^{t} (- 1 \cdot 1)$

Step 4.1.3.6

Simplify the expression.

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Step 4.1.3.6.1

Multiply $- 1$ by $1$ .

$(4 - t) \cdot 4^{t} \ln (4) + 4^{t} \cdot - 1$

Step 4.1.3.6.2

Move $- 1$ to the left of $4^{t}$ .

$(4 - t) \cdot 4^{t} \ln (4) - 1 \cdot 4^{t}$

Step 4.1.3.6.3

Rewrite $- 1 \cdot 4^{t}$ as $- 4^{t}$ .

$(4 - t) \cdot 4^{t} \ln (4) - 4^{t}$

Step 4.1.4

Simplify.

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Step 4.1.4.1

Apply the distributive property.

$(4 \cdot 4^{t} - t \cdot 4^{t}) \ln (4) - 4^{t}$

Step 4.1.4.2

Apply the distributive property.

$4 \cdot 4^{t} \ln (4) - t \cdot 4^{t} \ln (4) - 4^{t}$

Step 4.1.4.3

Multiply $4$ by $4^{t}$ .

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Step 4.1.4.3.1

Raise $4$ to the power of $1$ .

$4^{1} \cdot 4^{t} \ln (4) - t \cdot 4^{t} \ln (4) - 4^{t}$

Step 4.1.4.3.2

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

$4^{1 + t} \ln (4) - t \cdot 4^{t} \ln (4) - 4^{t}$

Step 4.1.4.4

Reorder terms.

$f' (t) = 4^{1 + t} \ln (4) - 4^{t} t \ln (4) - 4^{t}$

Step 4.2

The first derivative of $g (t)$ with respect to $t$ is $4^{1 + t} \ln (4) - 4^{t} t \ln (4) - 4^{t}$ .

$4^{1 + t} \ln (4) - 4^{t} t \ln (4) - 4^{t}$

Step 5

Set the first derivative equal to $0$ then solve the equation $4^{1 + t} \ln (4) - 4^{t} t \ln (4) - 4^{t} = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$4^{1 + t} \ln (4) - 4^{t} t \ln (4) - 4^{t} = 0$

Step 5.2

Reorder factors in $4^{1 + t} \ln (4) - 4^{t} t \ln (4) - 4^{t}$ .

$4^{1 + t} \ln (4) - t \cdot 4^{t} \ln (4) - 4^{t} = 0$

Step 5.3

Graph each side of the equation. The solution is the x-value of the point of intersection.

$t \approx 3.27865247$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

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 7

Critical points to evaluate.

$t = 3.27865247$

Step 8

Evaluate the second derivative at $t = 3.27865247$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$4^{1 + 3.27865247} \ln^{2} (4) - 2^{1 + 2 (3.27865247)} \ln (4) - 3.27865247 \cdot (4^{3.27865247} \ln^{2} (4))$

Step 9

Evaluate the second derivative.

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Step 9.1

Remove parentheses.

$4^{1 + 3.27865247} \ln^{2} (4) - 2^{1 + 2 (3.27865247)} \ln (4) - 3.27865247 \cdot (4^{3.27865247} \ln^{2} (4))$

Step 9.2

Simplify each term.

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Step 9.2.1

Add $1$ and $3.27865247$ .

$4^{4.27865247} \ln^{2} (4) - 2^{1 + 2 (3.27865247)} \ln (4) - 3.27865247 \cdot (4^{3.27865247} \ln^{2} (4))$

Step 9.2.2

Raise $4$ to the power of $4.27865247$ .

$376.70854776 \ln^{2} (4) - 2^{1 + 2 (3.27865247)} \ln (4) - 3.27865247 \cdot (4^{3.27865247} \ln^{2} (4))$

Step 9.2.3

Multiply $376.70854776$ by $\ln^{2} (4)$ .

$723.96302856 - 2^{1 + 2 (3.27865247)} \ln (4) - 3.27865247 \cdot (4^{3.27865247} \ln^{2} (4))$

Step 9.2.4

Multiply $2$ by $3.27865247$ .

$723.96302856 - 2^{1 + 6.55730495} \ln (4) - 3.27865247 \cdot (4^{3.27865247} \ln^{2} (4))$

Step 9.2.5

Add $1$ and $6.55730495$ .

$723.96302856 - 2^{7.55730495} \ln (4) - 3.27865247 \cdot (4^{3.27865247} \ln^{2} (4))$

Step 9.2.6

Raise $2$ to the power of $7.55730495$ .

$723.96302856 - 1 \cdot 188.35427388 \ln (4) - 3.27865247 \cdot (4^{3.27865247} \ln^{2} (4))$

Step 9.2.7

Multiply $- 1$ by $188.35427388$ .

$723.96302856 - 188.35427388 \ln (4) - 3.27865247 \cdot (4^{3.27865247} \ln^{2} (4))$

Step 9.2.8

Multiply $- 188.35427388$ by $\ln (4)$ .

$723.96302856 - 261.11446777 - 3.27865247 \cdot (4^{3.27865247} \ln^{2} (4))$

Step 9.2.9

Raise $4$ to the power of $3.27865247$ .

$723.96302856 - 261.11446777 - 3.27865247 \cdot (94.17713694 \ln^{2} (4))$

Step 9.2.10

Multiply $94.17713694$ by $\ln^{2} (4)$ .

$723.96302856 - 261.11446777 - 3.27865247 \cdot 180.99075714$

Step 9.2.11

Multiply $- 3.27865247$ by $180.99075714$ .

$723.96302856 - 261.11446777 - 593.40579467$

Step 9.3

Simplify by subtracting numbers.

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Step 9.3.1

Subtract $261.11446777$ from $723.96302856$ .

$462.84856078 - 593.40579467$

Step 9.3.2

Subtract $593.40579467$ from $462.84856078$ .

$- 130.55723388$

Step 10

$t = 3.27865247$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$t = 3.27865247$ is a local maximum

Step 11

Find the y-value when $t = 3.27865247$ .

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Step 11.1

Replace the variable $t$ with $3.27865247$ in the expression.

$g (3.27865247) = (4 - (3.27865247)) \cdot 4^{3.27865247}$

Step 11.2

Simplify the result.

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Step 11.2.1

Multiply $- 1$ by $3.27865247$ .

$g (3.27865247) = (4 - 3.27865247) \cdot 4^{3.27865247}$

Step 11.2.2

Subtract $3.27865247$ from $4$ .

$g (3.27865247) = 0.72134752 \cdot 4^{3.27865247}$

Step 11.2.3

Raise $4$ to the power of $3.27865247$ .

$g (3.27865247) = 0.72134752 \cdot 94.17713694$

Step 11.2.4

Multiply $0.72134752$ by $94.17713694$ .

$g (3.27865247) = 67.93444421$

Step 11.2.5

The final answer is $67.93444421$ .

$y = 67.93444421$

Step 12

These are the local extrema for $g (t) = (4 - t) \cdot 4^{t}$ .

$(3.27865247, 67.93444421)$ is a local maxima

Step 13