Find the critical points and the interval on which the given function is increasing or decreasing, and apply the First Derivative Test to each critical point
f(x)= (7/4)x^4+(28/3)x^3+(-63/2)x^2-252x
critical points in increasing order:
c1=? min or max?
c2=? min or max?
c3=? min or max?
y' = 7x^3+28x^2-63x-252
= 7(x+4)(x+3)(x-3)
Now things should be easy.
You know y is increasing when y' > 0...
To find the critical points of the given function f(x) = (7/4)x^4 + (28/3)x^3 + (-63/2)x^2 - 252x, we need to find the values of x where the derivative of the function equals zero or is undefined. The critical points are potential turning points where the function may have a minimum or maximum.
1. Find the derivative of the function f(x):
f'(x) = 4(7/4)x^3 + 3(28/3)x^2 + 2(-63/2)x - 252
Simplify:
f'(x) = 7x^3 + 28x^2 - 63x - 252
2. Set the derivative equal to zero and solve for x to find the critical points:
7x^3 + 28x^2 - 63x - 252 = 0
One way to solve this equation is by factoring out the common factor (x - 2):
(x - 2)(7x^2 + 42x + 126) = 0
Setting each factor equal to zero gives:
x - 2 = 0 and 7x^2 + 42x + 126 = 0
Solving the first equation, we find:
x = 2
To solve the second equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For the equation: 7x^2 + 42x + 126 = 0, a = 7, b = 42, and c = 126.
Substituting into the quadratic formula, we have:
x = (-42 ± √(42^2 - 4(7)(126))) / (2(7))
x = (-42 ± √(1764 - 3528)) / 14
x = (-42 ± √(-1764)) / 14
Since the discriminant (√(-1764)) is negative, there are no real solutions for x in this case. This means there are no additional critical points.
3. So, the critical points of the function are:
c1 = 2
Now, let's determine if each critical point is a minimum or maximum using the First Derivative Test:
- Substitute a value less than c1 (e.g., x = 0) into the derivative f'(x). If f'(x) is positive, then c1 is a local minimum. If f'(x) is negative, then c1 is a local maximum.
- Substitute a value greater than c1 (e.g., x = 3) into the derivative f'(x). If f'(x) is positive, then c1 is a local maximum. If f'(x) is negative, then c1 is a local minimum.
Calculating the values:
f'(0) = 7(0)^3 + 28(0)^2 - 63(0) - 252 = -252
f'(3) = 7(3)^3 + 28(3)^2 - 63(3) - 252 = 441 + 252 - 189 - 252 = 252
Since f'(0) is negative and f'(3) is positive, we can conclude that c1 = 2 is a local minimum.
Therefore,
c1 = 2 is a local minimum.