Question

How do you find and classify local maxima, local minima, and all critical points of $f\left( x \right)={{x}^{3}}+{{x}^{2}}-4x-4$?

Answer 1

In this problem, we have to find the local maxima, local minima, and all critical points of $f\left( x \right)={{x}^{3}}+{{x}^{2}}-4x-4$. We know that the critical points occur when the first derivative vanishes. i.e. when $f'\left( x \right)=0$ . We can find the local maxima and minima by substituting the critical points in the second derivative to check for the points to be maxima and minima.

Complete step by step solution:
We know that the given differential equation is,
$f\left( x \right)={{x}^{3}}+{{x}^{2}}-4x-4$…….. (1)
We can find the first derivative of the equation (1), we get
$\Rightarrow f'\left( x \right)=3{{x}^{2}}+2x-4$
We know that the critical points occur when the first derivative vanishes. i.e. when $f'\left( x \right)=0$

& \Rightarrow f'\left( x \right)=0 \\\ & \Rightarrow 3{{x}^{2}}+2x-4=0 \\\ \end{aligned}$$ Now we can use the quadratic formula to solve for x, which is the critical point. The quadratic formula for the general equation $$a{{x}^{2}}+bx+c=0$$ is, $$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$$ We can compare the general equation and the above equation, we get a = 3, b = 2, c = -4. We can substitute the values in the quadratic formula, we get $$\begin{aligned} & \Rightarrow x=\dfrac{-2\pm \sqrt{4-4\left( 3 \right)\left( -4 \right)}}{2\left( 3 \right)} \\\ & \Rightarrow x=\dfrac{-2\pm 2\sqrt{13}}{2\left( 3 \right)} \\\ & \Rightarrow x=\dfrac{-1\pm \sqrt{13}}{3} \\\ \end{aligned}$$ $$\Rightarrow x=\dfrac{-1+\sqrt{13}}{3},\dfrac{-1-\sqrt{13}}{3}$$. We can use the calculator to find the exact value, we get The critical points are x = 0.868, -1.535. We can now find the second derivative to find the local maxima and minima. $$\Rightarrow f''\left( x \right)=6x+2$$ We can now substitute the value of x = 0.868 in the above step, we get $$\Rightarrow f''\left( 0.868 \right)=6\left( 0.868 \right)+2=7.2>0$$ Hence the given function is minimum at x = 0.868. We can now substitute the value of x = -1.535 in the above step, we get $$\Rightarrow f''\left( -1.535 \right)=6\left( -1.535 \right)+2=-7.2<0$$ Hence the given function is maximum at x = -1.535. Therefore, the critical points are x = 0.868, -1.535. The given function is minimum at x = 0.868. The function is maximum at x = -1.535. **Note:** Students make mistakes while finding the value of local maxima and minima. We should remember that when $$f''\left( x \right)>0$$ , then the point is local minima and when $$f''\left( x \right)<0$$, then the point is local maxima. We should remember that, we should substitute the x value in the second derivative to get the maxima and minima.