Find Critical Points
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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 domain
Hence, 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 -
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Question 2 of 2
2. Question
Find the critical points of the curve `f(x)=3x^2+12x-5` 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+12x-5` `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+12x-5` `f(``-2``)` `=` $$3(\color{#007DDC}{-2}^2)+12(\color{#007DDC}{-2})-5$$ Substitute `x=-2` `=` `3(4)-24-5` `=` `12-24-5` `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+12x-5` `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+12x-5` `f(``-4``)` `=` $$3\color{#D800AD}{(-4)}^2+12\color{#D800AD}{(-4)}-5$$ `=` `3(16)-48-5` `=` `48-48-5` `=` `-5` A boundary point is at `(-4,-5)``f(x)` `=` `3x^2+12x-5` `f(``4``)` `=` $$3\color{#e65021}{(4)}^2+12\color{#e65021}{(4)}-5$$ `=` `3(16)+48-5` `=` `48+48-5` `=` `91` A boundary point is at `(4,91)`
Form the curve by connecting the points, then identify critical points
The 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 -