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Vilka [71]
3 years ago
7

Exponential function

Mathematics
2 answers:
Alina [70]3 years ago
8 0
F(x) = -4 * 2^(1-x)
f(x) = -4 * 2^1 / 2^x
f(x) = -8 / 2^x
f(x) = -8 (1/ 2)^x

answer is B. 
f(x) = -8 (1/ 2)^x

Romashka [77]3 years ago
6 0
We can break the term 2^{1-x} apart using the following law of exponents:

a^{x+y}=a^xa^y

Applying that, we find that

2^{1-x}=2^{1+(-x)}=2^1\cdot2^{-x}=2\cdot \frac{1}{2^x}

Substituting that into our original function, we have

f(x)=-4\cdot2^{1-x}\\
f(x)=-4\cdot2\cdot \frac{1}{2^x}

Which we can rewrite in the form f(x)=ab^x as

f(x)=-8\cdot\big( \frac{1}{2}\big)^x
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(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.

(b)  two eigen values of 'k' = 1, -1

for k = 1, eigen vector is \left[\begin{array}{c}0\\1\end{array}\right]

for k = -1 eigen vector is \left[\begin{array}{c}1\\0\end{array}\right]

See the figure for the graph:

(a) for any (x, y) ∈ R² the reflection of (x, y) over the y - axis is ( -x, y )

∴ x → -x hence '-1' is the eigen value.

∴ y → y hence '1' is the eigen value.

also, ( 1, 0 ) → -1 ( 1, 0 ) so ( 1, 0 ) is the eigen vector for '-1'.

( 0, 1 ) → 1 ( 0, 1 ) so ( 0, 1 ) is the eigen vector for '1'.

(b) ∵ T(x, y) = (-x, y)

T(x) = -x = (-1)(x) + 0(y)

T(y) =  y = 0(x) + 1(y)

Matrix Representation of T = \left[\begin{array}{cc}-1&0\\0&1\end{array}\right]

now, eigen value of 'T'

T - kI =  \left[\begin{array}{cc}-1-k&0\\0&1-k\end{array}\right]

after solving the determinant,

we get two eigen values of 'k' = 1, -1

for k = 1, eigen vector is \left[\begin{array}{c}0\\1\end{array}\right]

for k = -1 eigen vector is \left[\begin{array}{c}1\\0\end{array}\right]

Hence,

(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.

(b)  two eigen values of 'k' = 1, -1

for k = 1, eigen vector is \left[\begin{array}{c}0\\1\end{array}\right]

for k = -1 eigen vector is \left[\begin{array}{c}1\\0\end{array}\right]

Learn more about " Matrix and Eigen Values, Vector " from here: brainly.com/question/13050052

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