It is given that a differentiable function f(x)=4x^3+kx^2-36x-15 (k is a constant) is decreasing on -3/2<x<2 and increasing on 2<x<5. Find the value of k and the turning point(s) of the curve y=f(x).
you know that
f'(x) has a root at x=2.
It is negative for -3/2 < x < 2
It is positive for 2<x<5
Nothing is said about any domain outside (-3/2,5)
But, since you want a cubic f(x), f'(x) must be positive for x < -3/2, so it has another root at x = -3/2
f'(x) = a(2x+3)(x-2) = a(2x^2 - x - 6)
now we can get somewhere
f(x) = a(2/3 x^3 - 1/2 x^2 - 6x) + c
or, changing a and c (which are arbitrary) and clearing fractions,
f(x) = a(4x^3 - 3x^2 - 36x) + c
Now, we want f(x) as described above, so a = 1 and
f(x) = 4x^3 - 3x^2 - 36x - 15
To find the value of k and the turning point(s) of the curve y=f(x), we need to use the properties of differentiation. Here's how you can do it step by step:
1. Differentiate the function f(x) with respect to x to find its derivative, f'(x).
f'(x) = 12x^2 + 2kx - 36
2. Since the function f(x) is decreasing on -3/2 < x < 2, the derivative f'(x) must be negative in that interval.
Set f'(x) < 0 and solve for x:
12x^2 + 2kx - 36 < 0
Note: The discriminant of this quadratic equation should be positive since the parabola opens upwards.
3. Similarly, since the function f(x) is increasing on 2 < x < 5, the derivative f'(x) must be positive in that interval.
Set f'(x) > 0 and solve for x:
12x^2 + 2kx - 36 > 0
4. The solutions to the quadratic inequalities will give the range of x-values for which the function is decreasing and increasing, respectively.
5. The turning point(s) of the curve y=f(x) occur where the derivative f'(x) changes sign. In other words, where f'(x) = 0.
Set f'(x) = 12x^2 + 2kx - 36 = 0 and solve for x to find the x-coordinate(s) of the turning point(s).
6. Once you have found the value(s) of x, substitute them back into the original function f(x) to find the corresponding y-coordinate(s) of the turning point(s).
By following these steps, you can find the value of k and the turning point(s) of the curve y=f(x).