Calculus I - Derivatives of Exponential and Logarithm Functions

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Section 3.6 : Derivatives of Exponential and Logarithm Functions

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9. Determine if $\displaystyle V\left( t \right) = \frac{t}{{{{\bf{e}}^t}}}$ is increasing or decreasing at the following points.

$t = - 4$
$t = 0$
$t = 10$

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$t = - 4$ Show Solution

We know that the derivative of the function will give us the rate of change for the function and so we’ll need that.

$V'\left( t \right) = \frac{{\left( 1 \right){{\bf{e}}^t} - t\left( {{{\bf{e}}^t}} \right)}}{{{{\left( {{{\bf{e}}^t}} \right)}^2}}} = \frac{{{{\bf{e}}^t} - t{{\bf{e}}^t}}}{{{{\left( {{{\bf{e}}^t}} \right)}^2}}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{1 - t}}{{{{\bf{e}}^t}}}}}$

Now, all we need to do is evaluate the derivative at the point in question. So,

$V'\left( { - 4} \right) = \frac{5}{{{{\bf{e}}^{ - 4}}}} = 272.991 > 0$

$V'\left( { - 4} \right) > 0$ and so the function must be increasing at $t = - 4$ .

$t = 0$ Show Solution

We found the derivative of the function in the first part so here all we need to do is the evaluation.

$V'\left( 0 \right) = \frac{1}{{{{\bf{e}}^0}}} = 1 > 0$

$V'\left( 0 \right) > 0$ and so the function must be increasing at $t = 0$ .

$t = 10$ Show Solution

We found the derivative of the function in the first part so here all we need to do is the evaluation.

$V'\left( {10} \right) = \frac{{ - 9}}{{{{\bf{e}}^{10}}}} = - 0.0004086 < 0$

$V'\left( {10} \right) < 0$ and so the function must be decreasing at $t = 10$ .