The derivative of a function f is f'(x) = x^2 - 8/x: Then at x = 2, f has
A. an infection point B. a relative maximum C. a vertical tangent
D. a discontinuity E. a relative minimum
how do you test for all of these?
f'(2) = 4 - 8/2 = 0
f"(x) = 2x+8/x^2
f"(2) = 4 + 8/4 = 6
so, f'(2)=0 and f"(2)>0, means there is a local minimum there.
Recall that an inflection point is where f" = 0 (an infection point is just sick!)
a discontinuity is where f is undefined.
a vertical tangent is where f is defined but f' is undefined
thanks. so how would u see if f is undefined? how do you get f from f'?
when dealing with rational functions, the value is undefined when the denominator is zero.
So, f' and f" are undefined at x=0
f is the anti-derivative of f'. So,
f(x) = 1/3 x^3 - 8ln(x) + c
If you take the derivative of that, you'll get f' as given. Note that the value of c does not matter, since the derivative of a constant is zero.
To test for each of the given possibilities at x = 2, let's go through them one by one:
A. Infection point: An infection point occurs when the sign of the derivative changes from positive to negative or from negative to positive as x passes through a certain point. To test for this, we need to check the sign of f'(x) around x = 2. Evaluate f'(x) for values slightly less than 2 and values slightly greater than 2. If the sign changes from positive to negative (or vice versa), then there is an infection point. In this case, f'(x) = x^2 - 8/x. For x slightly less than 2 (e.g., x = 1.9), f'(x) will be negative since 1.9^2 is smaller than 8/1.9. For x slightly greater than 2 (e.g., x = 2.1), f'(x) will also be negative since 2.1^2 is larger than 8/2.1. Since the sign of f'(x) does not change around x = 2, there is no infection point. Therefore, option A is not correct.
B. Relative maximum: A relative maximum occurs when the derivative changes from positive to negative as x passes through a certain point. To check for a relative maximum, we need to examine the sign of the derivative around x = 2, similar to the previous case. If the sign changes from positive to negative, we have a relative maximum. However, as we observed earlier, the sign of f'(x) remains negative both slightly less than and slightly greater than 2. Therefore, there is no relative maximum at x = 2. Thus, option B is not correct.
C. Vertical tangent: A vertical tangent occurs when the derivative approaches positive or negative infinity as x approaches a specific value. To determine if there is a vertical tangent at x = 2, we need to evaluate the limit of f'(x) as x approaches 2. The limit of f'(x) as x approaches 2 is not infinite (positive or negative), but rather a finite value since both x^2 and 8/x are finite as x approaches 2. Hence, there is no vertical tangent at x = 2, so option C is not correct.
D. Discontinuity: A function has a discontinuity at a point when it is not defined or has different values from the right and left. To check for a discontinuity at x = 2, we need to evaluate the function f(x) at x = 2 and examine its behavior on both sides. In this case, the original function f(x) is not given, only its derivative f'(x) = x^2 - 8/x. The derivative alone does not provide information about continuity or discontinuity at x = 2. Therefore, we cannot determine if there is a discontinuity based solely on the derivative. Option D cannot be confirmed or ruled out with the given information.
E. Relative minimum: A relative minimum occurs when the derivative changes from negative to positive as x passes through a certain point. This is similar to testing for a relative maximum. Checking the sign of f'(x) around x = 2, we can observe that f'(x) changes from negative to positive as we move from x slightly less than 2 to x slightly greater than 2. Therefore, there is a relative minimum at x = 2. Hence, option E is correct.
In conclusion, at x = 2, function f has a relative minimum (option E).