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Find a, b,c, and d, such that the cubic f(x)=ax^3+bx^2+cx+d has a relative maximum at (-7, 163); has a relative minimum at (5, -125); and has a point of inflection at (-1, 19).

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The cubic function with the desired properties is:

f(x) = (-3/4)x³ - (9/4)x² - (113/4)x + 63/4

To find the values of a, b, c, and d, we can use the given information about the relative maximum, relative minimum, and point of inflection.

Relative Maximum:

The point (-7, 163) is a relative maximum. At this point, the derivative of the cubic function is equal to zero. Taking the derivative of the cubic function, we have:

f'(x) = 3ax² + 2bx + c

Setting x = -7 and f'(-7) = 0, we get:

49a - 14b + c = 0

Relative Minimum:

The point (5, -125) is a relative minimum. At this point, the derivative of the cubic function is equal to zero. Taking the derivative of the cubic function, we have:

f'(x) = 3ax² + 2bx + c

Setting x = 5 and f'(5) = 0, we get:

75a + 10b + c = 0

Point of Inflection:

The point (-1, 19) is a point of inflection. At this point, the second derivative of the cubic function changes sign. Taking the second derivative of the cubic function, we have:

f''(x) = 6ax + 2b

Setting x = -1, we get:

-6a + 2b = 0

Solving the system of equations formed by the above three equations, we can find the values of a, b, c, and d.

49a - 14b + c = 0

75a + 10b + c = 0

-6a + 2b = 0

Solving these equations, we find:

a = -3/4

b = -9/4

c = -113/4

d = 63/4

Therefore, the cubic function with the desired properties is:

f(x) = (-3/4)x³ - (9/4)x² - (113/4)x + 63/4

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