Answer :
25. The inflection point for f(x) = x³ + 30x² is (-10, -1000).
26. The inflection point for f(x) = x³ - 24x² is (8, -512).
To find the x and y coordinates of all inflection points for the given functions, we need to determine the points where the concavity of the functions changes. For the function f(x) = x³ + 30x², we find the inflection points by analyzing the second derivative. For the function f(x) = x³ - 24x², we also find the inflection points by examining the second derivative.
To find the inflection points of a function, we look for points where the concavity changes. This occurs when the second derivative changes sign. Therefore, we need to calculate the second derivative for each function.
25. For the function f(x) = x³ + 30x², we start by finding the first derivative:
f'(x) = 3x² + 60x
Next, we find the second derivative by differentiating f'(x):
f''(x) = 6x + 60
Now, we set f''(x) = 0 and solve for x to find the potential inflection points:
6x + 60 = 0
6x = -60
x = -10
So, the potential inflection point for f(x) = x³ + 30x² is at x = -10.
To determine the concavity at this point, we evaluate the second derivative for a value slightly greater and slightly smaller than x = -10. For example, at x = -11, f''(-11) = 6(-11) + 60 = -6, and at x = -9, f''(-9) = 6(-9) + 60 = 6. Since the sign changes from negative to positive as we move from left to right around x = -10, we can conclude that there is an inflection point at x = -10.
To find the corresponding y-coordinate, we substitute x = -10 back into the original function:
f(-10) = (-10)³ + 30(-10)² = -1000
Therefore, the inflection point for f(x) = x³ + 30x² is (-10, -1000).
26. For the function f(x) = x³ - 24x², we follow the same steps. First, we find the first derivative:
f'(x) = 3x² - 48x
Then, we find the second derivative:
f''(x) = 6x - 48
Setting f''(x) = 0, we solve for x to find the potential inflection points:
6x - 48 = 0
6x = 48
x = 8
The potential inflection point for f(x) = x³ - 24x² is at x = 8.
Evaluating the second derivative for values slightly greater and slightly smaller than x = 8, we find that f''(7) = -6 and f''(9) = 6. The sign change from negative to positive around x = 8 indicates an inflection point.
To find the y-coordinate, we substitute x = 8 back into the original function:
f(8) = (8)³ - 24(8)² = -512
Therefore, the inflection point for f(x) = x³ - 24x² is (8, -512).
In summary, for f(x) = x³ + 30x², the inflection point is (-10, -1000), and for f(x) = x³ - 24x², the inflection point is (8, -512).
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