How many mins/max are there in the equation x^3-4x^2-7x+10?
I said that there are 2 because if it is to the 3rd exponent, it increases then decreases then increases again when graphed.
Is my statement correct?
Thank You
To determine the number of minima and maxima of the equation x^3 - 4x^2 - 7x + 10, you need to understand the concept of critical points and the behavior of the graph.
The critical points are the values of x where the derivative of the function is either zero or undefined. To find them, take the derivative of the equation:
f'(x) = 3x^2 - 8x - 7
Now, set f'(x) equal to zero and solve for x:
3x^2 - 8x - 7 = 0
Using factoring or the quadratic formula, you can find that the solutions are x = -1 and x = 7/3.
These are the critical points of the equation, and they divide the number line into three intervals: (-∞, -1), (-1, 7/3), and (7/3, ∞).
Now, evaluate the function at points within each interval to determine the behavior of the graph:
1. For x < -1, choose a value, let's say x = -2. Calculate f(-2) = (-2)^3 - 4(-2)^2 - 7(-2) + 10. The result is -16 + 16 + 14 + 10 = 24.
2. For -1 < x < 7/3, choose a value, let's say x = 0. Calculate f(0) = 0^3 - 4(0)^2 - 7(0) + 10. The result is 10.
3. For x > 7/3, choose a value, let's say x = 2. Calculate f(2) = 2^3 - 4(2)^2 - 7(2) + 10. The result is 8 - 16 - 14 + 10 = -12.
From the calculations, you can see that the graph of the equation decreases from (-∞, -1) to a minimum at x = -1, then increases from there to a maximum at x = 7/3, and finally decreases again. This behavior indicates that there is one minimum and one maximum in the equation.
Therefore, your statement that there are two minima and maxima is incorrect. The correct answer is that there is one minimum and one maximum in the equation x^3 - 4x^2 - 7x + 10.