Question
h(x) =x^3+ax^2+bx-2 has a stationary point at (1,2) find h(x)
Answers
GPT 3.5
To find the equation of the function h(x) given that it has a stationary point at (1,2), we first need to determine the values of a and b.
A stationary point on a curve occurs at a point where the gradient of the curve is equal to zero. This means that the derivative of the function at that point is zero.
Given that the function h(x) = x^3 + ax^2 + bx - 2, the derivative of h(x) is given by:
h'(x) = 3x^2 + 2ax + b
Since the function has a stationary point at (1,2), we can set h'(1) = 0:
3(1)^2 + 2a(1) + b = 0
3 + 2a + b = 0
b = -2a - 3
Also, since the coordinates of the stationary point are (1,2), we can substitute these values into the function h(x):
2 = 1^3 + a(1)^2 + b(1) - 2
2 = 1 + a + b - 2
2 = a + b - 1
a + b = 3
Substitute b = -2a - 3 into a + b = 3:
a - 2a - 3 = 3
-a - 3 = 3
-a = 6
a = -6
Now that we have determined the value of a, we can substitute it back into b = -2a - 3:
b = -2(-6) - 3
b = 12 - 3
b = 9
Therefore, the equation of the function h(x) is:
h(x) = x^3 - 6x^2 + 9x - 2
A stationary point on a curve occurs at a point where the gradient of the curve is equal to zero. This means that the derivative of the function at that point is zero.
Given that the function h(x) = x^3 + ax^2 + bx - 2, the derivative of h(x) is given by:
h'(x) = 3x^2 + 2ax + b
Since the function has a stationary point at (1,2), we can set h'(1) = 0:
3(1)^2 + 2a(1) + b = 0
3 + 2a + b = 0
b = -2a - 3
Also, since the coordinates of the stationary point are (1,2), we can substitute these values into the function h(x):
2 = 1^3 + a(1)^2 + b(1) - 2
2 = 1 + a + b - 2
2 = a + b - 1
a + b = 3
Substitute b = -2a - 3 into a + b = 3:
a - 2a - 3 = 3
-a - 3 = 3
-a = 6
a = -6
Now that we have determined the value of a, we can substitute it back into b = -2a - 3:
b = -2(-6) - 3
b = 12 - 3
b = 9
Therefore, the equation of the function h(x) is:
h(x) = x^3 - 6x^2 + 9x - 2