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Critical Points (Maximum and Minimum Values)Critical Points (Maximum and Minimum Values)
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Question 1 of 2
1. Question
Find the critical points in the curve below:Enter the corresponding letter for the point
`(a)` Global Minimum: (A, a)`(b)` Global Maximum: (F, f)`(c)` Local Minimum: (D, d)`(d)` Local Maximum: (B, b)
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Critical points in a curve include the Global Minimum and Maximum (also called Absolute Minimum and Maximum) and Local Minimum and Maximum (also called Relative Minimum and Maximum)`(a)` Find the Global Minimum from the given curve.A global minimum has the lowest function value for all `x` in the domainHence, the Global Minimum is Point A.`(b)` Find the Global Maximum from the given curve.A global maximum has the highest function value for all `x` in the domain
[highlight only point F in green]Hence, the Global Minimum is Point F.`(c)` Find the Local Minimum from the given curve.A local minimum is a stationary or turning point that has the lowest function value for all `x` in a small interval around it. This is not applicable for endpoints or boundary points
[highlight only point D in green]Hence, the Local Minimum is Point D.`(d)` Find the Local Maximum from the given curve.A local maximum is a stationary or turning point that has the highest function value for all `x` in a small interval around it. This is not applicable for endpoints or boundary points
[highlight only point B in green]Hence, the Local Minimum is Point B.`(a)` Global Minimum: A`(b)` Global Maximum: F`(c)` Local Minimum: D`(d)` Local Maximum: B 

Question 2 of 2
2. Question
Find the critical points of the curve `f(x)=3x^2+12x5` under the domain `4≤x≤4`Write coordinates in the format “(a,b)”
Write `0` if the answer does not exist
`(a)` Global Minimum: ((2,17))`(b)` Global Maximum: ((4,91))`(c)` Local Minimum: ((2,17))`(d)` Local Maximum: (0)
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Critical points in a curve include the Global Minimum and Maximum (also called Absolute Minimum and Maximum) and Local Minimum and Maximum (also called Relative Minimum and Maximum)A point that satisfies `f'(x)=0` is called a Stationary PointFirst, find the stationary point of the curve by equating its first derivative to `0``f(x)` `=` `3x^2+12x5` `f'(x)` `=` `6x+12=0` Equate `f'(x)` to `0` `6(x+2)=0` Factor out `6` `x=2` This means that a stationary point exists when `x=2`Find the corresponding `y` value of the stationary point by solving for `f(``2``)``f(x)` `=` `3x^2+12x5` `f(``2``)` `=` $$3(\color{#007DDC}{2}^2)+12(\color{#007DDC}{2})5$$ Substitute `x=2` `=` `3(4)245` `=` `12245` `y` `=` `17` Hence, a stationary point is at `(2,17)`Next, confirm if this stationary point is a minimum or maximum by solving for `f”(x)``f(x)` `=` `3x^2+12x5` `f'(x)` `=` `6x+12` `f”(x)` `=` `6` This value is positive, which means the stationary point `(2,17)` is a minimum pointNow, find the two boundary points by solving for `f(``4``)` and `f(``4``)` (from `4≤x≤4`)`f(x)` `=` `3x^2+12x5` `f(``4``)` `=` $$3\color{#D800AD}{(4)}^2+12\color{#D800AD}{(4)}5$$ `=` `3(16)485` `=` `48485` `=` `5` A boundary point is at `(4,5)``f(x)` `=` `3x^2+12x5` `f(``4``)` `=` $$3\color{#e65021}{(4)}^2+12\color{#e65021}{(4)}5$$ `=` `3(16)+485` `=` `48+485` `=` `91` A boundary point is at `(4,91)`Form the curve by connecting the points, then identify critical pointsThe Global Minimum has the lowest function value for all `x` in the domainHence, it is the point `(2,17)`The Global Maximum has the highest function value for all `x` in the domainHence, it is the point `(4,91)`The Local Minimum is a stationary or turning point that has the lowest function value for all `x` in a small interval around it. This is not applicable for endpoints or boundary pointsHence, it is the point `(2,17)`The Local Maximum is a stationary or turning point that has the highest function value for all `x` in a small interval around it. This is not applicable for endpoints or boundary pointsHence, there is no local maximum`(a)` Global Minimum: `(2,17)``(b)` Global Maximum: `(4,91)``(c)` Local Minimum: `(2,17)``(d)` Local Maximum: does not exist 