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Section 3.6 : Derivatives of Exponential and Logarithm Functions
9. Determine if V(t)=tet is increasing or decreasing at the following points.
- t=−4
- t=0
- t=10
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a t=−4 Show SolutionWe 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 > 0V'\left( { - 4} \right) > 0 and so the function must be increasing at t = - 4.
b 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 > 0V'\left( 0 \right) > 0 and so the function must be increasing at t = 0.
c 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 < 0V'\left( {10} \right) < 0 and so the function must be decreasing at t = 10.