Question

How do you use the second derivative test to find where the function $f\left( x \right) = \dfrac{{5{e^x}}}{{5{e^x} + 6}}$ is concave up, concave down, and inflection points?

Answer 1

Hint : Second derivative of a function is the derivative of the derivative of the function. Second derivative test is generally used to locate the maxima and minima and the function and also to check for the concavity of the function. The critical points are calculated where the first derivative is zero and to check further for concavity we check whether the second derivative is positive or negative.

Complete step by step solution:
We have been given the function $f\left( x \right) = \dfrac{{5{e^x}}}{{5{e^x} + 6}}$ .
We have to check where the function is concave up, concave down and also find the inflection points.
First we find the first derivative of the given function.
$f'\left( x \right) = \dfrac{{d\left( {\dfrac{{5{e^x}}}{{5{e^x} + 6}}} \right)}}{{dx}} = \dfrac{{\left( {5{e^x} + 6} \right)\left( {5{e^x}} \right) - \left( {5{e^x}} \right)\left( {5{e^x}} \right)}}{{{{\left( {5{e^x} + 6} \right)}^2}}} = \dfrac{{30{e^x}}}{{{{\left( {5{e^x} + 6} \right)}^2}}}$
We can see that the function is not zero for any value of $x$ and it approaches $0$ when $x \to - \infty$ .
We can find the second derivative of the function as,
$f''\left( x \right) = \dfrac{{d\left( {\dfrac{{30{e^x}}}{{{{\left( {5{e^x} + 6} \right)}^2}}}} \right)}}{{dx}} = \dfrac{{{{\left( {5{e^x} + 6} \right)}^2}\left( {30{e^x}} \right) - \left( {30{e^x}} \right).2\left( {5{e^x} + 6} \right).\left( {5{e^x}} \right)}}{{{{\left( {5{e^x} + 6} \right)}^4}}} \\\ = \dfrac{{750{e^{3x}} + 1800{e^{2x}} + 1080{e^x} - 1500{e^{3x}} - 1800{e^{2x}}}}{{{{\left( {5{e^x} + 6} \right)}^4}}} \\\ = \dfrac{{ - 750{e^{3x}} + 1080{e^x}}}{{{{\left( {5{e^x} + 6} \right)}^4}}} \;$
We will use the following condition to test for concavity.
If $f''\left( x \right) > 0$ then $f\left( x \right)$ is concave up at those points.
If $f''\left( x \right) < 0$ then $f\left( x \right)$ is concave down at those points.
If $f''\left( x \right) = 0$ then these points are inflection points.
For $f''\left( x \right) > 0$ , we have,
$\dfrac{{ - 750{e^{3x}} + 1080{e^x}}}{{{{\left( {5{e^x} + 6} \right)}^4}}} > 0 \\\ \Rightarrow - 750{e^{3x}} + 1080{e^x} > 0\;\;\;[\because denominator\;is\;always\;positive] \\\ \Rightarrow 750{e^{3x}} < 1080{e^x} \\\ \Rightarrow {e^{2x}} < \dfrac{{1080}}{{750}} \Rightarrow {e^{2x}} < \dfrac{{36}}{{25}} \Rightarrow {e^x} < \dfrac{6}{5} \Rightarrow x < \ln \left( {1.2} \right) \\\ \Rightarrow x < 0.18 \;$
Similarly we can have for $f''\left( x \right) < 0$ ,
$\dfrac{{ - 750{e^{3x}} + 1080{e^x}}}{{{{\left( {5{e^x} + 6} \right)}^4}}} < 0 \\\ \Rightarrow - 750{e^{3x}} + 1080{e^x} < 0 \\\ \Rightarrow 750{e^{3x}} > 1080{e^x} \\\ \Rightarrow {e^{2x}} > \dfrac{{1080}}{{750}} \Rightarrow {e^{2x}} > \dfrac{{36}}{{25}} \Rightarrow {e^x} > \dfrac{6}{5} \Rightarrow x > \ln \left( {1.2} \right) \\\ \Rightarrow x > 0.18 \;$
And for point of inflection we have $f''\left( x \right) = 0$ ,
$\dfrac{{ - 750{e^{3x}} + 1080{e^x}}}{{{{\left( {5{e^x} + 6} \right)}^4}}} = 0 \\\ \Rightarrow - 750{e^{3x}} + 1080{e^x} = 0 \\\ \Rightarrow 750{e^{3x}} = 1080{e^x} \\\ \Rightarrow {e^{2x}} = \dfrac{{1080}}{{750}} \Rightarrow {e^{2x}} = \dfrac{{36}}{{25}} \Rightarrow {e^x} = \dfrac{6}{5} \Rightarrow x = \ln \left( {1.2} \right) \\\ \Rightarrow x = 0.18 \;$
Thus, the function is concave up for $x < 0.18$ , concave down for $x > 0.18$ and the point of inflection is $x = 0.18$ .

Note : We found the second derivative as the derivative of the first derivative of the function. The function is concave up where the second derivative is positive, concave down where the second derivative is negative and the point of inflection is the point where the sign of the second derivative changes, i.e. where the second derivative is zero.