Question

How do you find the critical points and the open intervals where the function is increasing and decreasing for $y=x{{e}^{x\left( 2-3x \right)}}$?

Answer 1

Take natural log, i.e., log to the base e, on both the sides of the given function to simplify it. Now, differentiate both the sides with respect to the variable x, use the chain rule of differentiation to simplify the L.H.S. while using the product rule of differentiation to simplify the R.H.S. substitute $\dfrac{dy}{dx}=0$ and find the values of x to get the critical points. Now, substitute $\dfrac{dy}{dx}>0$ and $\dfrac{dy}{dx}<0$ to find the open intervals in which the function is increasing and decreasing respectively.

Complete step by step solution:
Here, we have been provided with the function $y=x{{e}^{x\left( 2-3x \right)}}$ and we are asked to determine the critical points and the open intervals where this function is increasing and decreasing.
(i) Now, critical points are the values of x where the function has its derivative equal to 0. So, we need to differentiate the given function and substitute it equal to 0.
$\because y=x{{e}^{x\left( 2-3x \right)}}$
Taking log to the base e on both the sides, we get,
$\Rightarrow \ln y=\ln \left( x{{e}^{x\left( 2-3x \right)}} \right)$
Using the property of log given as: - $\log \left( m\times n \right)=\log m+\log n$ and $\log {{a}^{m}}=m\log a$, we get,

& \Rightarrow \ln y=\ln x+\ln {{e}^{x\left( 2-3x \right)}} \\\ & \Rightarrow \ln y=\ln x+x\left( 2-3x \right)\ln e \\\ & \Rightarrow \ln y=\ln x+2x-3{{x}^{2}} \\\ \end{aligned}$$ Differentiating both the sides with respect to x and using the chain rule of derivative in the L.H.S., we get, $$\begin{aligned} & \Rightarrow \dfrac{d\left( \ln y \right)}{dx}=\dfrac{d\left( \ln x \right)}{dx}+\dfrac{d\left( 2x-3{{x}^{2}} \right)}{dx} \\\ & \Rightarrow \dfrac{d\left( \ln y \right)}{dy}\times \dfrac{dy}{dx}=\dfrac{d\left( \ln x \right)}{dx}+\dfrac{d\left( 2x-3{{x}^{2}} \right)}{dx} \\\ & \Rightarrow \dfrac{1}{y}\left( \dfrac{dy}{dx} \right)=\dfrac{1}{x}+2-6x \\\ \end{aligned}$$ $$\Rightarrow \dfrac{dy}{dx}=y\left( \dfrac{1+2x-6{{x}^{2}}}{x} \right)$$ $$\begin{aligned} & \Rightarrow \dfrac{dy}{dx}=x{{e}^{x\left( 2-3x \right)}}\left( \dfrac{1+2x-6{{x}^{2}}}{x} \right) \\\ & \Rightarrow \dfrac{dy}{dx}={{e}^{x\left( 2-3x \right)}}\left( 1+2x-6{{x}^{2}} \right) \\\ \end{aligned}$$ Substituting $$\dfrac{dy}{dx}=0$$, we get, $$\Rightarrow {{e}^{x\left( 2-3x \right)}}\left( 1+2x-6{{x}^{2}} \right)=0$$ Since, exponential function is always positive, therefore $${{e}^{x\left( 2-3x \right)}}$$ cannot be equal to 0, so we have, $$\begin{aligned} & \Rightarrow 1+2x-6{{x}^{2}}=0 \\\ & \Rightarrow 6{{x}^{2}}-2x-1=0 \\\ \end{aligned}$$ Using the quadratic formula given as: - $$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$$, where a = co – efficient of $${{x}^{2}}$$, b = co – efficient of x and c = constant term, we get, $$\begin{aligned} & \Rightarrow x=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 6 \right)\left( -1 \right)}}{2\times 6} \\\ & \Rightarrow x=\dfrac{2\pm \sqrt{28}}{12} \\\ & \Rightarrow x=\dfrac{2\pm 2\sqrt{7}}{12} \\\ & \Rightarrow x=\dfrac{1\pm \sqrt{7}}{6} \\\ \end{aligned}$$ Hence, the critical points of the given function are: - $$\dfrac{1+\sqrt{7}}{6}$$ and $$\dfrac{1-\sqrt{7}}{6}$$. (ii) Now, let us determine the intervals in which the function is increasing. We know that a function is increasing only when its derivative is greater than 0, so mathematically we must have, $$\begin{aligned} & \Rightarrow \dfrac{dy}{dx}>0 \\\ & \Rightarrow {{e}^{x\left( 2-3x \right)}}\left( 1+2x-6{{x}^{2}} \right)>0 \\\ \end{aligned}$$ Since, the exponential function $${{e}^{x\left( 2-3x \right)}}$$ will always be greater than 0, so we must have the quadratic expression $$\left( 1+2x-6{{x}^{2}} \right)$$ greater than 0 to get the conditions satisfied. $$\begin{aligned} & \Rightarrow 1+2x-6{{x}^{2}}>0 \\\ & \Rightarrow 6{{x}^{2}}-2x-1<0 \\\ \end{aligned}$$ Since, we have found the roots of the equation $$6{{x}^{2}}-2x-1$$, so we can write it as: - $$\Rightarrow \left[ x-\left( \dfrac{1-\sqrt{7}}{6} \right) \right]\left[ x-\left( \dfrac{1+\sqrt{7}}{6} \right) \right]<0$$ Using the formula: - if $$\left( x-a \right)\left( x-b \right)<0$$ then $$a\[\begin{aligned} & \Rightarrow \dfrac{1-\sqrt{7}}{6} & \Rightarrow x\in \left( \dfrac{1-\sqrt{7}}{6},\dfrac{1+\sqrt{7}}{6} \right) \\\ \end{aligned}$$ (iii) At last, let us determine the intervals in which the function is decreasing. We know that a function is decreasing only when its derivative is less than 0, so mathematically we must have, $$\begin{aligned} & \Rightarrow \dfrac{dy}{dx}<0 \\\ & \Rightarrow {{e}^{x\left( 2-3x \right)}}\left( 1+2x-6{{x}^{2}} \right)<0 \\\ \end{aligned}$$ Since, the exponential function $${{e}^{x\left( 2-3x \right)}}$$ will always be greater than 0, so we must have the quadratic expression $$\left( 1+2x-6{{x}^{2}} \right)$$ less than 0 to get the conditions satisfied. $$\begin{aligned} & \Rightarrow 1+2x-6{{x}^{2}}<0 \\\ & \Rightarrow 6{{x}^{2}}-2x-1>0 \\\ & \Rightarrow \left[ x-\left( \dfrac{1-\sqrt{7}}{6} \right) \right]\left[ x-\left( \dfrac{1+\sqrt{7}}{6} \right) \right]>0 \\\ \end{aligned}$$ Using the formula: - if $$\left( x-a \right)\left( x-b \right)>0$$ then $$xb$$, where ‘a’ must be less than ‘b’, we get, $$\Rightarrow x<\dfrac{1-\sqrt{7}}{6}$$ or $$x>\dfrac{1+\sqrt{7}}{6}$$ $$\Rightarrow x\in \left( -\infty ,\dfrac{1-\sqrt{7}}{6} \right)\cup \left( \dfrac{1+\sqrt{7}}{6},\infty \right)$$ **Note:** One may wonder why we have taken log to the base e on both the sides at the initial step of the solution. The reason is nothing but to simply reduce the calculation we have done so. Although it is not necessary. You may note that if the function is undefined or even its derivative is undefined at any point then we have to include that point also in the set of critical points. You must remember the conditions for which the functions are increasing or decreasing otherwise it will be difficult to solve the above question.