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## AP®︎ Calculus BC (2017 edition)

### Course: AP®︎ Calculus BC (2017 edition)>Unit 6

Lesson 4: Finding relative extrema (first derivative test)

# 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?
• 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.
• 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
• 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
• How would you determine if an endpoint on a closed interval is a local minimum or local maximum?
• At , what exactly is the significance of Sal's "critical points"?
• You have to use the "critical points" to find the relative extrema.
• can discontinuous points be relative maxima or minima
• 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.
• Why 4/5 is a relative max? If the function increases from (0:4/5)? So 4/5 MUST BE GLOBAL MAX
• 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).
• 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?
• You can't subtract 1/2 and 5/16, so you change the denominator of 1/2 to 16, so you get 8/16.
• What about if the two critical numbers found are the same value? How would one go about finding the relative extrema?
• 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.
• 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.....
• 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! :)
• hou we can find the horizontal asemptot of a f(x) is fraction
• 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/b3. 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