Video: APCALC02AB-P1A-Q33-739105780826

Let 𝑓 and 𝑓′ be continuous and differentiable functions over the entire real numbers. The only critical points for 𝑓 are located at π‘₯ = βˆ’2 and π‘₯ = 7. If 𝑓″(π‘₯) = 4 βˆ’ 2π‘₯, which of the following must be true? I) 𝑓 has a relative minimum at π‘₯ = βˆ’2. II) 𝑓 has a relative minimum at π‘₯ = 7. III) 𝑓 has a point of inflection at π‘₯ = 2.

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Video Transcript

Let 𝑓 and 𝑓 prime be continuous and differentiable functions over the entire real numbers. The only critical points for 𝑓 are located at π‘₯ equals negative two and π‘₯ equals seven. If 𝑓 double prime of π‘₯ is equal to four minus two π‘₯, which of the following must be true? I) 𝑓 has a relative minimum at π‘₯ equals negative two. II) 𝑓 has a relative minimum at π‘₯ equals seven. III) 𝑓 has a point of inflection at π‘₯ equals two.

We’ll take each statement in turn to determine whether or not it’s true. For the first statement, if 𝑓 has a relative minimum at π‘₯ equals negative two, then this means that the first derivative 𝑓 prime of negative two is equal to zero, which tells us that 𝑓 has a critical point at π‘₯ equals negative two because a relative minimum is an example of a critical point.

The question tells us that there is indeed a critical point for 𝑓 located at π‘₯ equals negative two. So this condition for 𝑓 to have a relative minimum at π‘₯ equals negative two is fulfilled. However, all we know so far is that this is a critical point. We don’t know whether it’s a relative minimum, a relative maximum, or perhaps a point of inflection. If the point is a relative minimum, then the second derivative 𝑓 double prime evaluated at π‘₯ equals negative two will be positive. We can, therefore, substitute negative two into the expression given for the second derivative 𝑓 double prime of π‘₯ to test this statement.

𝑓 double prime of negative two will be equal to four minus two times negative two. That’s four plus four which is equal to eight. And as eight is greater than zero, the second derivative is indeed positive when π‘₯ equals negative two. So we found that the function 𝑓 does have a critical point at π‘₯ equals negative two. And as the second derivative is positive here, the point has indeed a relative minimum. The first statement is therefore true.

We can test the second statement in the same way. If the function 𝑓 has a relative minimum at π‘₯ equals seven, then the first derivative will be equal to zero at this point. And 𝑓 will have a critical point at π‘₯ equals seven. We’re told in the question this is indeed true. Again, if the point is to be a relative minimum, then the second derivative evaluated when π‘₯ equals seven must be positive.

Substituting seven into the function 𝑓 double prime of π‘₯ gives four minus 14 which is equal to negative 10. And this is less than zero. This tells us that whilst the function 𝑓 does have a critical point at π‘₯ equals seven, it is in fact a relative maximum, not a relative minimum. So the second statement is false.

The final statement concerns points of inflection. Now, points of inflection can also be critical points of a function. But they don’t have to be. We know that there isn’t a critical point for 𝑓 at π‘₯ equals two. We’re told that in the question because the only critical points are at π‘₯ equals negative two and π‘₯ equals seven. But this doesn’t mean that 𝑓 can’t have a point of inflection then. If 𝑓 does have a point of inflection at π‘₯ equals two, then it’ll be the case that the second derivative 𝑓 double prime evaluated at this point will be equal to zero.

Substituting two into our expression for 𝑓 double prime of π‘₯ then gives four minus two times two which is four minus four which is equal to zero. If the second derivative of a function at a point is equal to zero but the point isn’t a critical point, then this tells us that that point is a point of inflection. So we can conclude that statement three is also true.

To answer the question then, we can conclude that only statements one and three are true.

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