Question

Locate the critical points of the following functions. Then use the Second Derivative Test to determine (if possible) whether they correspond to local maxima or local minima. $$f(x)=e^{x}\left(x^{2}-7 x-12\right)$$

Short answer

Question: Determine the critical points of the function $$f(x)=e^x(x^2-7x-12)$$ and indicate whether they are local maxima or minima using the Second Derivative Test. Answer: The critical points $$x_1=\frac{5-\sqrt{97}}{2}$$ and $$x_2=\frac{5+\sqrt{97}}{2}$$ represent a local minimum and a local maximum, respectively.

Step-by-step solution

Compute the first derivative

Using the product rule, where $$(fg)'=f'g+fg'$$, calculate the first derivative of the given function:$$f(x)=e^x(x^2-7x-12)$$.$$f'(x)=\frac{d}{dx}\left[e^{x}\left(x^{2}-7 x-12\right)\right]=e^{x}\left(x^{2}-7 x-12\right)'+e^{x}'\left(x^{2}-7 x-12\right)=e^{x}(2x-7)+e^{x}(x^{2}-7x-12)$$

Find critical points

Critical points can be found by setting the first derivative equal to zero and solving for x:$$0=e^{x}(2x-7)+e^{x}(x^{2}-7x-12)$$.Factor out $$e^{x}$$:$$0=e^{x}\left((2x-7)+(x^{2}-7x-12)\right)$$.Therefore, $$0=x^2-5x-19$$, which we can solve using the quadratic formula: $$x=\frac{-(-5)\pm\sqrt{(-5)^{2}-4\cdot 1\cdot(-19)}}{2\cdot 1}$$.$$x=\frac{5\pm\sqrt{97}}{2}$$.We have two critical points: $$x_{1}=\frac{5-\sqrt{97}}{2}$$ and $$x_{2}=\frac{5+\sqrt{97}}{2}$$.

Compute the second derivative

Differentiate the first derivative again using the product rule:$$f''(x)=\frac{d}{dx}\left[e^{x}(2x-7)+e^{x}(x^{2}-7x-12)\right]=(2x-7)e^{x}+(x^2-7x-12)e^x+(2)e^{x}+(x^2-7x-12)e^x$$.$$f''(x)=e^{x}\left(2x-7+2+x^2-7x-12+x^2-7x-12\right)$$.$$f''(x)=e^{x}\left(2x^2-14x-2\right)$$.

Apply the Second Derivative Test

Evaluate the second derivative at each critical point to determine if it represents a local maximum or minimum: For $$x_{1}=\frac{5-\sqrt{97}}{2}$$:$$f''\left(\frac{5-\sqrt{97}}{2}\right)=e^{\frac{5-\sqrt{97}}{2}}\left(2\left(\frac{5-\sqrt{97}}{2}\right)^{2}-14\left(\frac{5-\sqrt{97}}{2}\right)-2\right)$$.It turns out that this expression is positive, meaning that $$x_1$$ corresponds to a local minimum. For $$x_{2}=\frac{5+\sqrt{97}}{2}$$:$$f''\left(\frac{5+\sqrt{97}}{2}\right)=e^{\frac{5+\sqrt{97}}{2}}\left(2\left(\frac{5+\sqrt{97}}{2}\right)^{2}-14\left(\frac{5+\sqrt{97}}{2}\right)-2\right)$$.This expression is negative, meaning that $$x_2$$ corresponds to a local maximum. #Conclusion# The critical points $$x_1=\frac{5-\sqrt{97}}{2}$$ and $$x_2=\frac{5+\sqrt{97}}{2}$$ represent a local minimum and a local maximum, respectively.

Key concepts

First Derivative Test

The First Derivative Test is a powerful tool used to determine where a function has local maxima and minima. By taking the derivative of the function and setting it to zero, the critical points are found. These points are where the function's rate of change switches direction, indicating a potential peak or valley.

For our function, we first calculated the derivative, which revealed the critical points. It is important to check each critical point to see if the function changes from increasing to decreasing, or vice versa, which helps us classify it as a local maximum or minimum. Keep in mind though, that this test doesn't say anything about global extrema or inflection points; for that you would need further analysis.

Second Derivative Test

The Second Derivative Test takes the analysis a step further by evaluating the concavity at the critical points. After finding the second derivative of the function, we plug in the critical points to see whether the function is concave up (indicating a local minimum) or concave down (indicating a local maximum).

In the given exercise, the test showed us that one critical point was a local minimum and the other was a local maximum, by checking whether the second derivative was positive or negative, respectively. This test is especially helpful when the First Derivative Test is inconclusive.

Product Rule Differentiation

The Product Rule is vital when differentiating functions that are the product of two other functions. It states that the derivative of a product is the derivative of the first function times the second function plus the first function times the derivative of the second function.In this exercise, we applied the Product Rule to differentiate the function:

Using this rule, we were able to find the first derivative efficiently, which was necessary for locating the critical points and for the Second Derivative Test.

Quadratic Formula

When simplifying the first derivative and setting it equal to zero, we often obtain a quadratic equation. To find the roots of this equation, we employ the Quadratic Formula:

The solution gives us the critical points of the function, which are essential in determining where potential maxima and minima occur. In our function, the quadratic formula was used to find the exact critical points, which we then used in the Second Derivative Test.