Since f(x)=√(x-3) has an absolute minimum of 0 does its critical point exist?
I'm really confused because someone mentioned that f(x)=√(x+3) does not have a critical point.
Does that mean that if there is an end point there, there can't be a critical point?
f(x) has an absolute minimum of 0 because of the definition of the square root.
Since we cannot take the square root of a negative number,
x-3 ≥ 0
x ≥ 3
and f(3) = 0
and f(x) only starts at x = 3
If you sketch y = √(x-3) , there are no points below the x-axis nor to the left of (3,0) and (3,0) is the minimum point.
It depends how your text or your instructor has defined "critical" point.
I'd say the critical point exists, since the usual definition that y'=0 or is undefined.
In either case, I think that (3,0) is the critical point.
Would -3,0 be the critical point for f(x)=√(x+3)?
To determine if a function has a critical point, we need to understand what a critical point is. A critical point occurs where the derivative of a function is either zero or undefined. In other words, it's a point where the slope of the function is either horizontal (zero derivative) or vertical (undefined derivative).
Let's analyze the function f(x) = √(x-3). To find the critical points, we first need to compute its derivative f'(x).
Using the chain rule, we differentiate f(x) = √(x-3) as follows:
f'(x) = (1/2) * (x-3)^(-1/2) * 1
Simplifying further, we have:
f'(x) = 1/(2√(x-3))
To check for critical points, we set f'(x) equal to zero and solve for x:
1/(2√(x-3)) = 0
However, we can observe that there is no value of x that makes the denominator zero. Therefore, the equation 1/(2√(x-3)) = 0 has no solutions.
Since there are no values of x for which the derivative is zero (or undefined), there are no critical points for the function f(x) = √(x-3).
Now, regarding the function f(x) = √(x+3), you mentioned someone said it doesn't have a critical point. That statement is incorrect. Let's find the derivative of f(x) = √(x+3):
Using the chain rule, we differentiate f(x) as follows:
f'(x) = (1/2) * (x+3)^(-1/2) * 1
Simplifying further, we have:
f'(x) = 1/(2√(x+3))
Similar to the previous case, we set f'(x) equal to zero to check for critical points:
1/(2√(x+3)) = 0
Again, there are no values of x that make the denominator zero. Therefore, the equation 1/(2√(x+3)) = 0 has no solutions.
So, the function f(x) = √(x+3) also does not have any critical points.
To answer your final question, the presence or absence of an end point does not directly relate to whether a function has a critical point. The critical point is solely determined by the behavior of the function's derivative. In either case, we have shown that both f(x) = √(x-3) and f(x) = √(x+3) do not have any critical points.