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3 years ago

F(x)=(x-6)e^-3x find the interval in which f(x) is increasing, the intervals on which f(x) is decreasing, and the local extrema

1 answer:

5 0

$\bf f(x)=(x-6)e^{-3x}\\\\
-----------------------------\\\\
\cfrac{dy}{dx}=1\cdot e^{-3x}+(x-6)-3e^{-3x}\implies \cfrac{dy}{dx}=e^{-3x}[1-3(x-6)]
\\\\\\
\cfrac{dy}{dx}=e^{-3x}(19-3x)\implies \cfrac{dy}{dx}=\cfrac{19-3x}{e^{3x}}$

set the derivative to 0, solve for "x" to get any critical points
keep in mind, setting the denominator to 0, also gives us critical points, however, in this case, the denominator will never be 0, so... no critical points from there

there's only 1 critical point anyway, and do a first-derivative test on it, check a number before it and after it, to see what sign the derivative has, and thus, whether the graph is going up or down, to check for any extrema

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Question is in picture! answer asap! will give brainliest!

nataly862011 [7]

Answer:solve for x

-3x+5y=-15

Add -5y to both sides

-3x+5y+-5y=-15+-5y

-3x=-5y-15

Divide both sides by -3

-3x/-3=-5y-15/-3

x= 5/3 y +5

I hope that's help!

Step-by-step explanation: plz give brainlist

8 0

3 years ago

Please dont ignore, Need help!!! Use the law of sines/cosines to find..

Ket [755]

Answer:

16. Angle C is approximately 13.0 degrees.

17. The length of segment BC is approximately 45.0.

18. Angle B is approximately 26.0 degrees.

15. The length of segment DF "e" is approximately 12.9.

Step-by-step explanation:

By the law of sine, the sine of interior angles of a triangle are proportional to the length of the side opposite to that angle.

For triangle ABC:

$\sin{A} = \sin{103\textdegree{}}$ ,
The opposite side of angle A $a = BC = 26$ ,
The angle C is to be found, and
The length of the side opposite to angle C $c = AB = 6$ .

$\displaystyle \frac{\sin{C}}{\sin{A}} = \frac{c}{a}$ .

$\displaystyle \sin{C} = \frac{c}{a}\cdot \sin{A} = \frac{6}{26}\times \sin{103\textdegree}$ .

$\displaystyle C = \sin^{-1}{(\sin{C}}) = \sin^{-1}{\left(\frac{c}{a}\cdot \sin{A}\right)} = \sin^{-1}{\left(\frac{6}{26}\times \sin{103\textdegree}}\right)} = 13.0\textdegree{}$ .

Note that the inverse sine function here $\sin^{-1}()$ is also known as arcsin.

By the law of cosine,

$c^{2} = a^{2} + b^{2} - 2\;a\cdot b\cdot \cos{C}$ ,

where

$a$ , $b$ , and $c$ are the lengths of sides of triangle ABC, and
$\cos{C}$ is the cosine of angle C.

For triangle ABC:

$b = 21$ ,
$c = 30$ ,
The length of $a$ (segment BC) is to be found, and
The cosine of angle A is $\cos{123\textdegree}$ .

Therefore, replace C in the equation with A, and the law of cosine will become:

$a^{2} = b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}$ .

$\displaystyle \begin{aligned}a &= \sqrt{b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}}\\&=\sqrt{21^{2} + 30^{2} - 2\times 21\times 30 \times \cos{123\textdegree}}\\&=45.0 \end{aligned}$ .

For triangle ABC:

$a = 14$ ,
$b = 9$ ,
$c = 6$ , and
Angle B is to be found.

Start by finding the cosine of angle B. Apply the law of cosine.

$b^{2} = a^{2} + c^{2} - 2\;a\cdot c\cdot \cos{B}$ .

$\displaystyle \cos{B} = \frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}$ .

$\displaystyle B = \cos^{-1}{\left(\frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}\right)} = \cos^{-1}{\left(\frac{14^{2} + 6^{2} - 9^{2}}{2\times 14\times 6}\right)} = 26.0\textdegree$ .