Locate all critical points ( both types ) of f(x)=sqrt(-58-4x^2+32x)
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To locate the critical points of the function f(x) = sqrt(-58 - 4x^2 + 32x), we need to find the values of x that make the derivative of the function equal to zero or undefined.
Let's start by finding the derivative of f(x) and setting it equal to zero:
f(x) = sqrt(-58 - 4x^2 + 32x)
f'(x) = (1/2)(-4x + 32) / sqrt(-58 - 4x^2 + 32x)
Now, set f'(x) = 0 and solve for x:
0 = (-4x + 32) / (2 * sqrt(-58 - 4x^2 + 32x))
0 = -4x + 32
Solve the equation:
4x = 32
x = 8
Thus, x = 8 is a potential critical point of the function.
Now, let's check if there are any critical points where the derivative is undefined. To find these points, we need to locate values of x where the denominator of f'(x) is equal to zero.
Notice that the denominator of f'(x) is 2 * sqrt(-58 - 4x^2 + 32x). We need to find values of x that make this equal to zero:
2 * sqrt(-58 - 4x^2 + 32x) = 0
To solve this equation, we divide both sides by 2:
sqrt(-58 - 4x^2 + 32x) = 0
Now, to simplify the equation, we square both sides:
-58 - 4x^2 + 32x = 0
Rearrange the equation:
4x^2 - 32x + 58 = 0
Now, solve this quadratic equation.
You can use the quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / (2a).
In this case, a = 4, b = -32, and c = 58. Plug in these values to calculate the roots.
x = (-(-32) ± sqrt((-32)^2 - 4 * 4 * 58)) / (2 * 4)
Simplify the equation:
x = (32 ± sqrt(1024 - 928)) / 8
x = (32 ± sqrt(96)) / 8
x = (32 ± 4sqrt(6)) / 8
x = (4 ± sqrt(6)) / 2
Thus, the critical points are x = 8, (4 + sqrt(6))/2, and (4 - sqrt(6))/2.