Which of these statements is true for f(x)=(1/2)^x ?

Mathematics

2 answers:

Anarel [89]3 years ago

5 0

Answer:

Step-by-step explanation:

We are given that a function

$f(x)=(\frac{1}{2})^x$

It can be write as

$y=f(x)=2^{-x}$

Taking log on both sides

$logy=-xlog2$

Differentiate w.r.t x

$\frac{1}{y}\frac{dy}{dx}=-log 2$

$\frac{d(logx)}{dx}=\frac{1}{x}$

$\frac{dy}{dx}=-ylog2=-2^{-x}log2$ for all x

When f'(x) <0 then function is decreasing.

Hence, given function is decreasing function.

Substitute x=0 then we get

$f(0)=1$ Because ( $a^0=1$ )

Therefore, y intercept is (0,1).

Domain pf function f(x)=R

Range of given function :( $0,\infty$ )

Hence, option C is true.

s2008m [1.1K]3 years ago

3 0

Answer:

The answer is c

Step-by-step explanation:

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(a) The parametric equations x = f(t) and y = g(t) give the coordinates of a point (x, y) = (f(t), g(t)) for appropriate values

Gnesinka [82]

Answer:

Step-by-step explanation:

a). Given a parametric equation, we are describing a set of coordinates based on the value of t. The variable t is called the parameter.

b) we have the following equations. x=t y=t^2, so in order for us to know where the object is at t=t' we must replace t with the specific value t'. Hence, when t=0 the object is at (0,0^2) = (0,0) (the origin). When t=6, the object is at (6,6^2) = (6,36).

c). To eliminate the parameter, we replace the parameter in one equation by using the second equation. Recall that we have that x=t. Then, by replacing in the second equation, we have the following

$y=t^2 = (x)^2 = x^2$