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3 years ago

Find the absolute extrema if they exist as well as all values of x where they occur for the function f(x)=-x/x^2+5

1 answer:

4 0

$\bf f(x)=\cfrac{x}{x^2+5}
\\\\\\
\cfrac{df}{dx}=\cfrac{(x^2+5)-2x^2}{(x^2+5)^2}\implies \cfrac{df}{dx}=\cfrac{5-x^2}{(x^2+5)^2}\impliedby 
\begin{array}{llll}
using\ the\\
quotient\ rule
\end{array}\\\\
-------------------------------\\\\
0=\cfrac{5-x^2}{(x^2+5)^2}\implies 0=5-x^2\implies x^2=5\implies x=\pm\sqrt{5}
\\\\\\
f(\sqrt{5})\approx 0.2236\impliedby \textit{only maximum, thus absolute maximum}
\\\\\\
f(-\sqrt{5})\approx -0.2236\impliedby \textit{only minimum, thus absolute minimum}$

we also get critical points when the denominator is 0, namely (x²+5)² = 0

however, this denominator, doesn't give us any critical points

critical points when the denominator is 0, are usually asymptotic or "cusps", where the derivative is not continuous, but has an extrema.

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3 years ago

Read 2 more answers

Suppose that the data for analysis includes the attributeage. Theagevalues for the datatuples are (in increasing order) 13, 15,

Bas_tet [7]

Answer:

a) $\bar X = \frac{\sum_{i=1}^{27} X_i }{27}= \frac{809}{27}=29.96$

$Median = 25$

b) $Mode = 25, 35$

Since 25 and 35 are repeated 4 times, so then the distribution would be bimodal.

c) $Midrange = \frac{70+13}{3}=41.5$

d) $Q_1 = \frac{20+21}{2} =20.5$

$Q_3 =\frac{35+35}{2}=35$

e) Min = 13 , Q1 = 20.5, Median=25, Q3= 35, Max = 70

f) Figura attached.

g) When we use a quantile plot is because we want to show the percentage or the fraction of values below or equal to an specified value for the distribution of the data.

By the other hand the quantile-quantile plot shows the quantiles of the distribution values against other selected distribution (specified, for example the normal distribution). If the points are on a straight line we assume that the data values fit very well to the hypothetical distribution selected.

Step-by-step explanation:

For this case w ehave the following dataset given:

13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25, 30,33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70.

Part a

The mean is calculated with the following formula:

$\bar X = \frac{\sum_{i=1}^{27} X_i }{27}= \frac{809}{27}=29.96$

The median on this case since we have 27 observations and that represent an even number would be the 14 position in the dataset ordered and we got:

$Median = 25$

Part b

The mode is the most repeated value on the dataset on this case would be:

$Mode = 25, 35$

Since 25 and 35 are repeated 4 times, so then the distribution would be bimodal.

Part c

The midrange is defined as:

$Midrange = \frac{Max+Min}{2}$

And if we replace we got:

$Midrange = \frac{70+13}{3}=41.5$

Part d

For the first quartile we need to work with the first 14 observations

13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25

And the Q1 would be the average between the position 7 and 8 from these values, and we got:

$Q_1 = \frac{20+21}{2} =20.5$

And for the third quartile Q3 we need to use the last 14 observations:

25, 30,33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70

And the Q3 would be the average between the position 7 and 8 from these values, and we got:

$Q_3 =\frac{35+35}{2}=35$

Part e

The five number summary for this case are:

Min = 13 , Q1 = 20.5, Median=25, Q3= 35, Max = 70

Part f

For this case we can use the following R code:

> x<-c(13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25, 30,33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70)

> boxplot(x,main="boxplot for the Data")

And the result is on the figure attached. We see that the dsitribution seems to be assymetric. Right skewed with the Median<Mean

Part g

When we use a quantile plot is because we want to show the percentage or the fraction of values below or equal to an specified value for the distribution of the data.