find the values for a, b, and c such that the function f(x)= x^3 + ax^2 + bx+ c
has a critical point at (1,5) and an inflection point at (2,3).
a= -6
b= 3
c= 7
i got those, but they're wrong. i'm not sure why. :/
f = x^3+ax^2+bx+c
f' = 3x^2+2ax+b
f'' = 6x+2a
for inflection, f''=0, so
6(2)+2a=0
a=-6
for critical point,
f'=0, so
3(1)+2(-6)(1)+b=0
b=9
f(x) = x^3 - 6x^2 + 9x + c
5=1-6+9+c
c=1
3=8-24+18+c
c=1
so,
f = x^3 - 6x^2 + 9x + 1
f' = 3x^2-12x+9 so f'(1) = 0
f'' = 6x-12 so f''(2)=0
To find the values for a, b, and c, we can use the information given about the critical and inflection points.
We know that the critical point occurs at x = 1, and the inflection point occurs at x = 2.
For a critical point to occur at (1,5), the derivative of the function must be equal to zero at x = 1. Let's find the derivative of f(x) and set it equal to zero:
f'(x) = 3x^2 + 2ax + b
Setting f'(1) = 0, we have:
3(1)^2 + 2a(1) + b = 0
3 + 2a + b = 0 -- (1)
Now let's consider the inflection point at (2,3). For an inflection point to occur, the second derivative of the function must change sign. Let's find the second derivative of f(x):
f''(x) = 6x + 2a
Since the inflection point occurs at x = 2, we have:
f''(2) = 6(2) + 2a = 0
12 + 2a = 0 -- (2)
We now have a system of two equations with two variables (a and b):
Equation (1): 3 + 2a + b = 0
Equation (2): 12 + 2a = 0
Solving this system of equations:
From Equation (2) we get: 2a = -12
a = -6
Substituting a = -6 into Equation (1) gives: 3 + 2(-6) + b = 0
3 - 12 + b = 0
b = 9
Therefore, the correct values for a, b, and c are:
a = -6, b = 9, and c could be any real number since it was not specified in the problem statement.
To find the values for a, b, and c such that the function f(x) = x^3 + ax^2 + bx + c has a critical point at (1,5) and an inflection point at (2,3), we can use the properties of critical points and inflection points.
To find the critical point, we need to take the derivative of the function and set it equal to zero. The critical point occurs when the derivative is zero.
First, let's find the derivative of f(x):
f'(x) = 3x^2 + 2ax + b
Now, let's find the values of a and b that satisfy the condition of having a critical point at (1,5). We substitute x = 1 and f'(1) = 0 into the derivative equation:
0 = 3(1)^2 + 2a(1) + b
0 = 3 + 2a + b
Now, let's find the equation that represents the relationship between a and b:
-2a - b = 3 -----> (Equation 1)
Next, we need to find the equation for the inflection point. An inflection point occurs when the second derivative changes sign. So, we need to find the second derivative of f(x) and set it equal to zero at x = 2.
The second derivative is found by taking the derivative of the derivative:
f''(x) = 6x + 2a
Substituting x = 2 and f''(2) = 0 into the second derivative equation, we have:
0 = 6(2) + 2a
0 = 12 + 2a
-2a = 12
a = -6
Substituting a = -6 into Equation 1, we can solve for b:
-2(-6) - b = 3
12 - b = 3
-b = 3 - 12
-b = -9
b = 9
So, the correct values for a and b are a = -6 and b = 9.
Now, we can substitute these values into f(x) to find c. Using the critical point equation again:
0 = 3 + 2a + b
0 = 3 + 2(-6) + 9
0 = 3 - 12 + 9
0 = 0
This means that our equation holds true regardless of the value of c. Therefore, the value of c can be any real number.
In conclusion, the correct values for a, b, and c are a = -6, b = 9, and c can be any real number.