In the triangle below, with right angle , suppose that and .

Find the minimum value if f(x) = xe^x over [-2,0]

SashulF [63]

Given function:

$f(x)=xe^x$

The minimum value of the function can be found by setting the first derivative of the function to zero.

$f^{\prime}(x)=xe^x+e^x$ $\begin{gathered} xe^x+e^x\text{ = 0} \\ e^x(x\text{ + 1) = 0} \end{gathered}$

Solving for x:

$\begin{gathered} x\text{ + 1 = 0} \\ x\text{ = -1} \end{gathered}$ $\begin{gathered} e^x\text{ = 0} \\ \text{Does not exist} \end{gathered}$

Substituting the value of x into the original function:

$\begin{gathered} f(x=1)=-1\times e^{^{-1}}_{} \\ =\text{ -}0.368 \end{gathered}$

Hence, the minimum value in the given range is (-1, -0.368)

6 0

1 year ago

SOMEONE HELP WITH THIS PLZZ!!

Natali5045456 [20]

Answer:

x=90 y=45

Step-by-step explanation:

Answer is C

y and 45 are vertical angles

x=90 (right angle)

3 0

2 years ago

Please help asap!! Due today!

Gwar [14]

Answer:

No for the first

Step-by-step explanation:

Sorry i do not know the second one bt the first one is No it is not a right angle

5 0

2 years ago

Which temperature is closest to 0 degrees Celcius (°C)?

sladkih [1.3K]

Answer:

1 degree celcius. it is only 1 degree away from 0 degrees celcius. hope this helps

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Determine the formula for the nth term of the sequence:<br>-2,1,7,25,79,...

rodikova [14]

A plausible guess might be that the sequence is formed by a degree-4* polynomial,

$x_n = a n^4 + b n^3 + c n^2 + d n + e$

From the given known values of the sequence, we have

$\begin{cases}a+b+c+d+e = -2 \\ 16 a + 8 b + 4 c + 2 d + e = 1 \\ 81 a + 27 b + 9 c + 3 d + e = 7 \\ 256 a + 64 b + 16 c + 4 d + e = 25 \\ 625 a + 125 b + 25 c + 5 d + e = 79\end{cases}$

Solving the system yields coefficients

$a=\dfrac58, b=-\dfrac{19}4, c=\dfrac{115}8, d = -\dfrac{65}4, e=4$

so that the n-th term in the sequence might be

$\displaystyle x_n = \boxed{\frac{5 n^4}{8}-\frac{19 n^3}{4}+\frac{115 n^2}{8}-\frac{65 n}{4}+4}$

Then the next few terms in the sequence could very well be

$\{-2, 1, 7, 25, 79, 208, 466, 922, 1660, 2779, \ldots\}$

It would be much easier to confirm this had the given sequence provided just one more term...

* Why degree-4? This rests on the assumption that the higher-order forward differences of $\{x_n\}$ eventually form a constant sequence. But we only have enough information to find one term in the sequence of 4th-order differences. Denote the k-th-order forward differences of $\{x_n\}$ by $\Delta^{k}\{x_n\}$ . Then

• 1st-order differences:

$\Delta\{x_n\} = \{1-(-2), 7-1, 25-7, 79-25,\ldots\} = \{3,6,18,54,\ldots\}$

• 2nd-order differences:

$\Delta^2\{x_n\} = \{6-3,18-6,54-18,\ldots\} = \{3,12,36,\ldots\}$

• 3rd-order differences:

$\Delta^3\{x_n\} = \{12-3, 36-12,\ldots\} = \{9,24,\ldots\}$

• 4th-order differences:

$\Delta^4\{x_n\} = \{24-9,\ldots\} = \{15,\ldots\}$

From here I made the assumption that $\Delta^4\{x_n\}$ is the constant sequence {15, 15, 15, …}. This implies $\Delta^3\{x_n\}$ forms an arithmetic/linear sequence, which implies $\Delta^2\{x_n\}$ forms a quadratic sequence, and so on up $\{x_n\}$ forming a quartic sequence. Then we can use the method of undetermined coefficients to find it.

5 0

2 years ago