Thread: Calculus

The function f(x)=x^3+ax^2+bx has a local minimum at x=3 and a point of inflection at x=-1. Find the values of a and b.

Originally Posted by SoYah

The function f(x)=x^3+ax^2+bx has a local minimum at x=3 and a point of inflection at x=-1. Find the values of a and b.

Differentiate it, and set the result equal to zero.



But you know that this equation has two solutions, x = 3, and x =-1.

Apply them





2 equations, 2 unknowns. Solve simultaneously.

If f(x)= x^3+ax^2+bx, then
f '(x)= 3x^2+2ax+b and
f "(x)= 6x+2a.

f(x) has an inflection point at x= -1, which means that f "(x)=0 when x=-1. So f "(x)= 6x+2a=0 when x=-1
6(-1)+2a=0
-6+2a=0
2a=6
a=3.

f(x) also has a minimum at x=3, which means that f '(x)=0 when x=3. So,
f '(x)= 3x^2+2ax+b=0 when x=3
3(3)^2+2a(3)+b=0
We already have the value for a, so we can substitute it as well.
27+6(3)+b=0
45+b=0
b=-45.

Originally Posted by CaitSydney22

If f(x)= x^3+ax^2+bx, then
f '(x)= 3x^2+2ax+b and
f "(x)= 6x+2a.

f(x) has an inflection point at x= -1, which means that f "(x)=0 when x=-1. So f "(x)= 6x+2a=0 when x=-1
6(-1)+2a=0
-6+2a=0
2a=6
a=3.

f(x) also has a minimum at x=3, which means that f '(x)=0 when x=3. So,
f '(x)= 3x^2+2ax+b=0 when x=3
3(3)^2+2a(3)+b=0
We already have the value for a, so we can substitute it as well.
27+6(3)+b=0
45+b=0
b=-45.

It isn't really necessary to consider f''(x). See my post above.

All you have to know is that the given x coordinates are the coordinates of stationary points, and hence are solutions to the equation f'(x) = 0. You have two of these points, and two unknown variables, so it is simply to formulate two equations which can be solved simultaneously to find the values of a and b.