Question – 1: The gradient of the graph of f(x) at any point is,ax^2-12x + 9, and the point,(2,-3),is an inflection point of f(x).
(a) Find the value of the real constant, a. Find the function f(x).
(b) Find the stationary points of the function, then specify the intervals on which the function is increasing and those on which the function is decreasing
(c) Find the intervals on which f(x) is concave up, and those on which it is concave down.
1. Integrate the gradient function
ax^2-12x + 9 to get f(x), except for an arbitary constant C.
f(x) = (ax^3)/3 -6x^2 +9x + C
If (2,-3) is an inflection point, the second derivative of f(x) must be zero there.
2ax -12 = 0 @ x=2
ax = 6
a = 3
Use f(x) and the statement that y=-3 at x=2 to solve for C
(b) Stationary points occur where the derivative of f(x) is zero.
(c) f(x) is concave down where f''(x) < 0 and concave up where f''(x) >0
To find the value of the constant "a" and the function f(x), we need to integrate the given gradient function.
(a) To find "a," we integrate the given gradient function with respect to x:
∫ (ax^2 - 12x + 9) dx = f(x)
Integrating each term individually, we get:
∫ ax^2 dx - ∫ 12x dx + ∫ 9 dx = f(x)
Taking the integral of each term:
(a * (x^3/3) - 12 * (x^2/2) + 9x) + C = f(x)
where C is the constant of integration.
Now, we know that the point (2, -3) is an inflection point, which means the second derivative of f(x) is zero at this point. Therefore, we need to find the second derivative of f(x) and set it equal to zero to solve for "a."
Differentiating f(x) once more, we get:
f''(x) = 2a - 12
Setting f''(2) = 0, we have:
2a - 12 = 0
2a = 12
a = 6
So, the value of the constant "a" is 6.
Now, substitute the value of "a" back into the integrated function:
f(x) = (6 * (x^3/3) - 12 * (x^2/2) + 9x) + C
(b) To find the stationary points, we need to find the derivative of f(x) and set it equal to zero:
f'(x) = 6x^2 - 12x + 9
Setting f'(x) = 0, we have:
6x^2 - 12x + 9 = 0
This is a quadratic equation. We can solve it using factoring, completing the square, or the quadratic formula.
(c) To find the intervals where f(x) is concave up or concave down, we need to determine the sign of the second derivative, f''(x).
Taking the derivative of f'(x), we get:
f''(x) = 12x - 12
To find the intervals where f(x) is concave up, we need to find where f''(x) > 0.
12x - 12 > 0
12x > 12
x > 1
So, the function f(x) is concave up for x > 1.
To find the intervals where f(x) is concave down, we need to find where f''(x) < 0.
12x - 12 < 0
12x < 12
x < 1
So, the function f(x) is concave down for x < 1.
These steps allow you to solve the problem and find the value of "a," the function f(x), the stationary points, and the intervals where the function is increasing or decreasing, as well as where the function is concave up or concave down.