find values for a, b, and c so 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).
I got a as -6, but I don't know what b and c are. For b and c I just have b + c= 10. I do not know how to proceed thanks
Inflection point at (2,3) means f"(2)=0
=>
f"(x)=6x+2a = 0 at x=2, or
a=-6(2)/2=-6
Since f(2)=3 (from given point (2,3) ), we conclude that:
f(2)=c+2b-16=3
or
2b+c=19 ....(1)
We also know that there is a critical point at (1,5), so
f'(1)=0, or
b=9
Substituting b=9 in (1) gives
b=9, c=1
Check if f(1)=5.... indeed.
So problem solved.
To find values for b and c that satisfy the given conditions, we can use the properties of critical points and inflection points.
1. Critical Point at (1,5):
A critical point occurs when the derivative of the function is zero. Let's take the derivative of f(x) to find the conditions for the critical point:
f(x) = x^3 + ax^2 + bx + c
Taking the derivative:
f'(x) = 3x^2 + 2ax + b
Since the critical point is at (1,5), we can substitute x=1 and f(x)=5 into the derivative equation:
5 = 3(1)^2 + 2a(1) + b
5 = 3 + 2a + b --> Equation 1
2. Inflection Point at (2,3):
An inflection point occurs when the second derivative changes sign. Let's take the second derivative of f(x):
f'(x) = 3x^2 + 2ax + b
Taking the derivative again:
f''(x) = 6x + 2a
Since the inflection point is at (2,3), we can substitute x=2 and f''(x)=0 into the second derivative equation:
0 = 6(2) + 2a
0 = 12 + 2a
-2a = 12
a = -6 --> Equation 2
Substituting the value of a from Equation 2 into Equation 1:
5 = 3 + 2(-6) + b
5 = 3 - 12 + b
5 = -9 + b
b = 14
Now that we have obtained the values of a and b, we can find c by using the equation b + c = 10:
14 + c = 10
c = -4
Therefore, the values for a, b, and c that satisfy the given conditions are:
a = -6
b = 14
c = -4
To find the values of b and c, we can use the given information about the critical point and inflection point.
1. Critical Point:
A critical point occurs when the derivative of a function is equal to zero. In this case, the derivative of f(x) is:
f'(x) = 3x^2 + 2ax + b
Since the critical point is at (1,5), we know that f'(1) = 0. Substituting x=1 into the derivative:
3(1)^2 + 2a(1) + b = 0
3 + 2a + b = 0
2. Inflection Point:
An inflection point occurs when the second derivative of a function changes sign. In this case, the second derivative of f(x) is:
f''(x) = 6x + 2a
Since the inflection point is at (2,3), we know that f''(2) = 0. Substituting x=2 into the second derivative:
6(2) + 2a = 0
12 + 2a = 0
2a = -12
a = -6
Now that we have found a = -6, let's substitute this value into the equations we derived earlier.
3 + 2(-6) + b = 0
-12 + b = 0
b = 12
And the equation b + c = 10 becomes:
12 + c = 10
c = -2
Therefore, the values of a, b, and c that satisfy the given conditions are:
a = -6, b = 12, c = -2.