Estimate the local minimum of $y=2 x^3+x^2-7 x-6$.
A. there is no local minimum
B. $(-1.25,0.41)$
C. $(2,0)$
D. $(1,-10)$

The local minimum of the function y = 2 x 3 + x 2 − 7 x − 6 is approximately at the point ( 1 , − 10 ) , determined through the first and second derivative tests. The critical points were found by setting the first derivative to zero, and using the second derivative confirmed the nature of these points. Thus, the chosen option is D: ( 1 , − 10 ) .
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Answered by Anonymous | 2025-07-04

Find the first derivative: y ′ = 6 x 2 + 2 x − 7 .
Find critical points by solving 6 x 2 + 2 x − 7 = 0 , resulting in x ≈ 0.926 and x ≈ − 1.259 .
Use the second derivative y ′′ = 12 x + 2 to test for local minimum ( 0"> y ′′ > 0 ) and maximum ( y ′′ < 0 ).
The local minimum is at approximately ( 1 , − 10 ) ​ .

Explanation

Problem Analysis We are given the function y = 2 x 3 + x 2 − 7 x − 6 and asked to estimate the local minimum. To do this, we will find the critical points by taking the first derivative, setting it equal to zero, and solving for x . Then, we will use the second derivative test to determine whether each critical point is a local minimum or local maximum.

Finding the First Derivative First, find the first derivative of the function: y ′ = d x d ​ ( 2 x 3 + x 2 − 7 x − 6 ) = 6 x 2 + 2 x − 7

Finding Critical Points Next, set the first derivative equal to zero and solve for x to find the critical points: 6 x 2 + 2 x − 7 = 0 We can use the quadratic formula to solve for x :
x = 2 a − b ± b 2 − 4 a c ​ ​ where a = 6 , b = 2 , and c = − 7 .
x = 2 ( 6 ) − 2 ± 2 2 − 4 ( 6 ) ( − 7 ) ​ ​ = 12 − 2 ± 4 + 168 ​ ​ = 12 − 2 ± 172 ​ ​ = 12 − 2 ± 2 43 ​ ​ = 6 − 1 ± 43 ​ ​ So the critical points are x = 6 − 1 + 43 ​ ​ and x = 6 − 1 − 43 ​ ​ .

Finding the Second Derivative Now, find the second derivative of the function: y ′′ = d x 2 d 2 ​ ( 2 x 3 + x 2 − 7 x − 6 ) = d x d ​ ( 6 x 2 + 2 x − 7 ) = 12 x + 2

Second Derivative Test Evaluate the second derivative at each critical point to determine whether it is a local minimum or maximum. For x = 6 − 1 + 43 ​ ​ :
0"> y ′′ ( 6 − 1 + 43 ​ ​ ) = 12 ( 6 − 1 + 43 ​ ​ ) + 2 = 2 ( − 1 + 43 ​ ) + 2 = − 2 + 2 43 ​ + 2 = 2 43 ​ > 0 Since the second derivative is positive, this critical point is a local minimum. For x = 6 − 1 − 43 ​ ​ :
y ′′ ( 6 − 1 − 43 ​ ​ ) = 12 ( 6 − 1 − 43 ​ ​ ) + 2 = 2 ( − 1 − 43 ​ ) + 2 = − 2 − 2 43 ​ + 2 = − 2 43 ​ < 0 Since the second derivative is negative, this critical point is a local maximum.

Finding the Coordinates of Local Minimum The local minimum occurs at x = 6 − 1 + 43 ​ ​ . We can approximate this value: x = 6 − 1 + 43 ​ ​ ≈ 6 − 1 + 6.557 ​ ≈ 6 5.557 ​ ≈ 0.926 Now, we evaluate the original function at this x -value to find the corresponding y -value: y ( 0.926 ) = 2 ( 0.926 ) 3 + ( 0.926 ) 2 − 7 ( 0.926 ) − 6 ≈ 2 ( 0.795 ) + 0.857 − 6.482 − 6 ≈ 1.59 + 0.857 − 6.482 − 6 ≈ − 10.035 So the local minimum is approximately at ( 0.926 , − 10.035 ) .


Similarly, the local maximum occurs at x = 6 − 1 − 43 ​ ​ . We can approximate this value: x = 6 − 1 − 43 ​ ​ ≈ 6 − 1 − 6.557 ​ ≈ 6 − 7.557 ​ ≈ − 1.259 Now, we evaluate the original function at this x -value to find the corresponding y -value: y ( − 1.259 ) = 2 ( − 1.259 ) 3 + ( − 1.259 ) 2 − 7 ( − 1.259 ) − 6 ≈ 2 ( − 1.994 ) + 1.585 + 8.813 − 6 ≈ − 3.988 + 1.585 + 8.813 − 6 ≈ 0.41 So the local maximum is approximately at ( − 1.259 , 0.41 ) .

Final Answer Comparing our results with the given options, the local minimum is approximately at ( 0.926 , − 10.035 ) , which is closest to option D ( 1 , − 10 ) . The local maximum is approximately at ( − 1.259 , 0.41 ) , which is close to option B ( − 1.25 , 0.41 ) . Therefore, the local minimum is at approximately ( 1 , − 10 ) .

Conclusion Based on the calculations performed using the derivative tests, the local minimum is estimated to be at ( 1 , − 10 ) .


Examples
Understanding local minima and maxima is crucial in various real-world applications. For instance, in engineering, minimizing the cost function of a project or maximizing the efficiency of a machine often involves finding the local minimum or maximum of a function. In economics, businesses aim to maximize profits, which can be achieved by identifying the local maximum of the profit function. Similarly, in physics, determining the point of minimum potential energy is essential for understanding the stability of a system. These concepts are fundamental in optimization problems across diverse fields, enabling informed decision-making and efficient resource allocation.
For example, consider a company trying to minimize the cost of production. The cost function might be represented as C ( x ) = 0.1 x 3 − 1.5 x 2 + 7 x + 50 , where x is the number of units produced. To find the minimum cost, we would take the derivative, C ′ ( x ) = 0.3 x 2 − 3 x + 7 , set it to zero, and solve for x . The second derivative, C ′′ ( x ) = 0.6 x − 3 , would then be used to confirm that the critical point is indeed a minimum. This process helps the company determine the optimal production level to minimize costs.

Answered by GinnyAnswer | 2025-07-04