Given T(2,5,4) and M(3,1,-4), find the ordered triple that represents TM. Then find the magnitude of TM

I have calculus problems that I need help with.

aleksklad [387]

a. Note that $f(x)=x^ne^{-2x}$ is continuous for all $x$ . If $f(x)$ attains a maximum at $x=3$ , then $f'(3) = 0$ . Compute the derivative of $f$ .

$f'(x) = nx^{n-1} e^{-2x} - 2x^n e^{-2x}$

Evaluate this at $x=3$ and solve for $n$ .

$n\cdot3^{n-1} e^{-6} - 2\cdot3^n e^{-6} = 0$

$n\cdot3^{n-1} = 2\cdot3^n$

$\dfrac n2 = \dfrac{3^n}{3^{n-1}}$

$\dfrac n2 = 3 \implies \boxed{n=6}$

To ensure that a maximum is reached for this value of $n$ , we need to check the sign of the second derivative at this critical point.

$f(x) = x^6 e^{-2x} \\\\ \implies f'(x) = 6x^5 e^{-2x} - 2x^6 e^{-2x} \\\\ \implies f''(x) = 30x^4 e^{-2x} - 24x^5 e^{-2x} + 4x^6 e^{-2x} \\\\ \implies f''(3) = -\dfrac{486}{e^6} < 0$

The second derivative at $x=3$ is negative, which indicate the function is concave downward, which in turn means that $f(3)$ is indeed a (local) maximum.

b. When $n=4$ , we have derivatives

$f(x) = x^4 e^{-2x} \\\\ \implies f'(x) = 4x^3 e^{-2x} - 2x^4 e^{-2x} \\\\ \implies f''(x) = 12x^2 e^{-2x} - 16x^3e^{-2x} + 4x^4e^{-2x}$

Inflection points can occur where the second derivative vanishes.

$12x^2 e^{-2x} - 16x^3 e^{-2x} + 4x^4 e^{-2x} = 0$

$12x^2 - 16x^3 + 4x^4 = 0$

$4x^2 (3 - 4x + x^2) = 0$

$4x^2 (x - 3) (x - 1) = 0$

Then we have three possible inflection points when $x=0$ , $x=1$ , or $x=3$ .

To decide which are actually inflection points, check the sign of $f''$ in each of the intervals $(-\infty,0)$ , $(0, 1)$ , $(1, 3)$ , and $(3,\infty)$ . It's enough to check the sign of any test value of $x$ from each interval.

$x\in(-\infty,0) \implies x = -1 \implies f''(-1) = 32e^2 > 0$

$x\in(0,1) \implies x = \dfrac12 \implies f''\left(\dfrac12\right) = \dfrac5{43} > 0$

$x\in(1,3) \implies x = 2 \implies f''(2) = -\dfrac{16}{e^4} < 0$

$x\in(3,\infty) \implies x = 4 \implies f''(4) = \dfrac{192}{e^8} > 0$

The sign of $f''$ changes to either side of $x=1$ and $x=3$ , but not $x=0$ . This means only $\boxed{x=1}$ and $\boxed{x=3}$ are inflection points.

4 0

1 year ago

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What's the answer ?<br><br><br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B245%20%2B%2044%7D%20" id="TexFormula1" title=" \sq

-Dominant- [34]

$\huge\mathfrak\pink{Answer...}$

+,- 17

$\huge\mathfrak\purple{hope \: it \: helps..}$

8 0

3 years ago

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What is 2x/3 - 4 =x ?

Inessa [10]

Step-by-step explanation:

2x/3 -4 = x

multiply both side by 3

2x - 12 = 3x

-x = 12

x = -12

3 0

3 years ago

PLEASE HELP WILL MARK BRAINLIEST YOU DONT HAVE TO EXPLAIN

Yakvenalex [24]

Answer:

the second option

Step-by-step explanation:

8 0

3 years ago

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For a standard normal distribution, if P (z less-than-or-equal-to a) = 0.7116, what is the value of P (z greater-than-or-equal-t

Irina-Kira [14]

Answer:equal

Step-by-step explanation:

The probability that a random variable is greater than or equal to z standard deviations from the mean in a standard normal distribution he P(z≤a)+P(z>a)=1, so P(z>a)=1-P(z≤a)=1-.7116=.2884

6 0

2 years ago