Yx -³ / 3x⁰ Please help!!

2 answers:

6 0

Hey, i’d love to help you but there seems to be a typo in your equation because a minus sign (or negative sign) can’t be raised to the power of 3.

Marat540 [252]2 years ago

4 0

I'll assume you meant yx^(-3) over 3x^0.

Rewrite this as:

-----------------

x^3*3*x^0

Note that x^3 * x^0 = x^(3+0) = x^3, so that we are left with

------- is the simplest possible answer

x^3

Note that x^0 = 1

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The function F(x)=log0.75^x is decreasing. True or false

Novay_Z [31]

We're given the following function:
$f(x)=log(.75^x)=log[ (\frac{3}{4}) ^{x}]=log( \frac{3^x}{4^x})$

In order to see if the function is decreasing we'll take its derivative. If $\frac{d}{dx}f(x)\ \textgreater \ 1$ the function is increasing, if $\frac{d}{dx}f(x)\ \ \textless \ \ 1$ the function is decreasing.

We take the derivate:
$\frac{d}{dx}[log( \frac{3^x}{4^x})]= \frac{4^x}{3^x}[ \frac{d}{dx} (\frac{3^x}{4^x})]=\frac{4^x}{3^x} \frac{4^x[ \frac{d}{dx} (3^x)]-3^x[ \frac{d}{dx} (4^x)]}{ 4^2^x}=$
$\frac{4^x}{3^x} \frac{4^x[3^xlog(3)]-3^x[4^x[4^xlog(4)]}{ 4^2^x}=log(3)-log(4)\ \textless \ 0$

Which implies the function is decreasing.

Another way to answer the problem (although less insightful) you can take any two real numbers $k$ and $q$ such that $k\ \textgreater \ q$ , then if $f(k)-f(q)\ \textgreater \ 0$ the function is increasing and if $f(k)-f(q)\ \textless \ 0$ the function is decreasing. You can verify the function is decreasing with any two numbers in the function's domain.