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Zarrin [17]
3 years ago
11

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

Mathematics
1 answer:
DIA [1.3K]3 years ago
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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Considere as seguintes afirmações: I. O inverso de 0,2 é 5. II. O triplo de \frac{2}{5} é \frac{6}{15}. III. A metade de 0,5 é \
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I e III

Step-by-step explanation:

Recebemos declarações na questão acima para considerar.

Declaração I. O inverso de 0,2 é 5.

Matematicamente, isso é escrito como:

1 / 0,2

0,2 = 2/10

Portanto: 1 ÷ 2/10 = 1 × 10/2

= 5

Afirmação 1 é verdadeira

Declaração II.

O triplo de 2/5 é 6/15.

Triplo de 2/5 = 2/5 + 2/5 + 2/5

= 2 + 2 + 2/5

= 6/5

= 1 1/5

Declaração II é falsa

III. A metade de 0,5 é 1/5

1/2 de 0,5

= 1/2 × 0,5 = 0,25

Convertendo em fração

= 0,25 = 25/100

= 1/4

Declaração III está correta

Portanto, as afirmações verdadeiras são I e III

8 0
2 years ago
A line has a slope of 5 and passes through the point
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Answer:

Step-by-step explanation:

The equation is mx+b=y.

M=slope (in this case 5)

x=x coordinates

b=y intercept

y=y coordinates

For the x and y coordinates, I'll use (2,0).

5(2)+_=0

10+_=0.

In this case, we only have one awnser, -10. Since 10+(-10) are the only ways to equal 0.

Hope this helps,

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