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### Course: AP®︎/College Calculus BC > Unit 5

Lesson 4: Using the first derivative test to find relative (local) extrema- Introduction to minimum and maximum points
- Finding relative extrema (first derivative test)
- Worked example: finding relative extrema
- Analyzing mistakes when finding extrema (example 1)
- Analyzing mistakes when finding extrema (example 2)
- Finding relative extrema (first derivative test)
- Relative minima & maxima
- Relative minima & maxima review

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# Worked example: finding relative extrema

Sal finds the relative maximum point of g(x)=x⁴-x⁵ by analyzing the intervals where its derivative, g', is negative or positive.

## Want to join the conversation?

- So, Sal keeps mentioning that if the derivative is undefined, then the point is a critical point. Does that mean that if the slope of the tangent line is vertical at that given point, does that make that point a critical point?(10 votes)
- If the tangent line to a point (x, y) is vertical, then the derivative (slope of the vertical line) is undefined, so the function has a critical point at (x, y).

A critical point occurs wherever the derivative is either 0 or undefined. All extrema of a function will occur at critical points, though not all critical points will end up being extrema.(23 votes)

- Ok, so if the the derivative of f(x) is a fraction, how woud I find the critical numbers? Do i set the numerator and denominator equal to 0?

ex) the derivative of f(x) = (-4x+4)/(x+4)^4(5 votes)- A critical point is a point where the derivative equals 0 or undefined so you would set the numerator equal to zero to find where f'(x)=0 and the denominator equal to 0 to find f'(x)=undefined(7 votes)

- How would you determine if an endpoint on a closed interval is a local minimum or local maximum?(4 votes)
- At1:44, what exactly is the significance of Sal's "critical points"?(4 votes)
- You have to use the "critical points" to find the relative extrema.(2 votes)

- can discontinuous points be relative maxima or minima(2 votes)
- Nope. Maxima and minima need to be finite numbers. So, a point where the function is undefined (either a hole discontinuity, asymptote etc) cannot be considered an extremum.(4 votes)

- Why 4/5 is a relative max? If the function increases from (0:4/5)? So 4/5 MUST BE GLOBAL MAX(3 votes)
- Global maximums are also relative maximums, so 4/5 is both a relative maximum and a global maximum. This is because the requirement for relative maximums is met where there is a global maximum (f' goes from f'>0 to f'<0).(2 votes)

- how did you get 8/16?? you were doing g'(1/2) and you got 1/2 - 5/16 and than became 8/16 - 5/16 how is this?(2 votes)
- You can't subtract 1/2 and 5/16, so you change the denominator of 1/2 to 16, so you get 8/16.(2 votes)

- What about if the two critical numbers found are the same value? How would one go about finding the relative extrema?(2 votes)
- If two critical points occur at the same x value then it is actually just one and the same critical point. So you only have to check that one point.(2 votes)

- It bothers me how he keeps giving us a graph with a drawing that has one of Xs having f'(x) = undefined, yet doesn't give us an equation that could result to the drawing of a graph that has a "pointy" extrema.....(0 votes)
- An absolute value function has a "pointy" extrema ("corner"). Also some piecewise defined functions can have cusps or corners as well resulting in an undefined derivative at that point.

Hope this helps! :)(8 votes)

- hou we can find the horizontal asemptot of a f(x) is fraction(2 votes)
- To find the horizontal asymptote of a rational of a function
`f(x)=p(x) where q(x) and p(x) are polynomials`

---

q(x)

we take the`lim f(x) and lim f(x)`

x->∞ x->-∞

1. But generally there are 3 cases, either the highest degree of p(x) is higher than q(x) which implies that`f(x) has no horizontal asymptote.`

2. both highest degrees of p(x) & q(x) are equal then if p(x)'s highest

degree is a and q(x)'s highest degree is b then the horizontal asymptote is a/b

3. if q(x)'s highest degree is higher than p(x)'s

, then the horizontal asymptote is 0.

Ex.find horizontal limit of f(x)=(x^2+2x-1)/(x^2-3).

1. if we use the cases then it is the case which they are equal and both coefficients are 1 thus the horizontal asymptote is 1.

second way. evaluate the limit.

1. lim as x->∞ of (x^2+2x-1)/(x^2-3)

2. lim as x->∞ of (x^2/x^2+2x/x^2-1/x^2)/(x^2/x^2-3/x^2)(divide the numerator and denominator by x^2)

3. lim as x->∞ of (x^2/x^2)/(x^2/x^2) (simplifying)

4. lim as x->∞ of 1 = 1 (by limit laws)

we can evaluate the other limit with the same process`*Hope this helps*(2 votes)

## Video transcript

- [Voiceover] So we have g(x) being equal to x to the fourth minus x to the fifth, and what we wanna do
without having to graph g, we want to figure out at what x values does g have a relative maximum? And just to remind us what's
going on in a relative maximum, so let me draw a hypothetical
function right over here, so a relative maximum is going to happen, so you can visually inspect this, okay that looks like a relative maximum. That's kind of a top of a
mountain or top of a hill, these all look like relative
maximum and what's in common? Well the graph, the function is going from increasing to decreasing
at each of those points. It's going from increasing to decreasing. Increasing to decreasing
at either of the points, or you could say that the first derivative is going from positive to negative. So if you look at this
interval right over here, g prime is greater than zero, and then over the next interval
when you're decreasing, g prime would be less than zero. So what we really need to think about is when does g prime, so let me see, relative, we care about
relative maximum point, and so that's essentially asking when does g prime go from positive to negative? From, from, I wrote fror. From g prime greater than zero, to g prime less than zero, and the values that we
could look at or the points are our critical points
and critical points are where g prime is either
zero or it is undefined. So let's think about it, where is g prime of x equal to zero? g prime of x is equal to zero when well let's just take g prime of x. We're gonna leverage the
power rule right here, four x to the third power,
four x to the third, minus five x to the fourth. Minus five x to the
fourth is equal to zero. Let's see, we can factor
out an x to the third. So we have x to the third times four minus five x is equal to zero. So this is going to happen
when x is equal to zero. Let me not skip steps. So this is going to happen when x to the third is equal to zero, or four minus five x is equal to zero. For x to the third equaling zero, that's only gonna happen
when x is equal to zero, and four minus five x equaling zero, we'll add five x to both sides, you get four is equal to five x, divide both sides by five,
you get 4/5 is equal to x. So here these are the two places, where our derivative is equal to zero. Now are there any places where
our derivative is undefined? Well our function right over here is just a straight up polynomial. Our derivative is another polynomial, it is defined for all real numbers. So these are our two critical points, or we could even say critical values. Now let's think of what g
prime is doing on either side of these critical values, and I'll draw a little number line here to help us visualize this, and so. So there we go, little
bit of a number line. Let's see we care about
zero and we care about 4/5. So let's say this is negative 1, this is zero, this is one, and so we have one critical point at, let me do this in magenta, we have one critical point
here at x equals zero, and then we have another critical point. I will do this at x equals 4/5. So 4/5 is right around there. So that is 4/5 and let's just think about what g prime is doing in these intervals, and these critical points
are the only places where g prime might switch
sides, switch signs. So let's first think about this, let me pick some colors
I haven't used yet. So let's think about the interval from negative infinity to zero. So this is the open interval
from negative infinity to zero, and we could just plug in a value, let's try negative one,
negative one is pretty straight forward to evaluate. So let's see you have four,
you're gonna have four times negative one to the third power. So that's gonna be four
times negative one, minus five times negative
one to the fourth power. So that's just gonna be one. So let's see this is going to
be negative four minus five. Which is negative nine,
so right over here, g prime is equal to negative nine, and so we know over this whole interval, since it's to the left
of this critical point, we know that g prime
of x is less than zero, and so our function itself is decreasing over this interval and
so we know we need to go from increasing to decreasing, so you can already say well
we can't go from increasing to decreasing at this critical point, because we're already
decreasing to the left of it, but anyway let's just think about what's happening at the other intervals. So in the interval between zero and 4/5, so that interval right over there, so it's between zero and 4/5, well let's just sample a number there. Let's say the number, I don't know, 1/2? Might be fairly straightforward. So we can evaluate g prime
of 1/2, g prime of 1/2, is equal to four times
1/2 to the third power. 1/2 to the third power is 1/8. So it's 4/8, or it's just 1/2 minus five times 1/2 to the fourth so
that's 5/16 minus 5/16 and so this is equal to 8/16 minus 5/16 which is equal to 3/16, but the important thing is
it's equal to a positive value. So in this blue interval right over there, and actually let me put
4/5 in a different color so we see that it's not
part of that interval. So in this light blue interval right here between zero and 4/5, g prime, g prime of x is greater than zero. So we know our function is increasing, and so let's see what's
happening to the right of this, and the easiest value to
try out would just be one. So let's try out x equals one. It's in that interval. So when x equals one, I'll just write g prime of one is equal
to four minus five. Four minus five, which
is equal to negative one. So g prime of x is less than zero, g prime of x is less than zero. So our function, so we could
say g is increasing here. It is decreasing, oh
sorry let me be careful, g is decreasing here. The function itself is decreasing cause our derivative is negative. Then our function is increasing here cause our derivative is positive, and then our function is decreasing here. So at what critical
point are we going from increasing to decreasing? Well we're doing that at x equals 4/5. So we have a relative
maximum at x equals 4/5. 4/5, if they said, "Well where do we have "a relative minimum point?" Well that's going to be
happen at x equals zero. We're going from decreasing to increasing, but we've answered their question of where do we find a
relative maximum point.