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Rzqust [24]
3 years ago
13

Find the domain and range of the exponential function h(x) = 125x.

Mathematics
1 answer:
kirill [66]3 years ago
4 0
H(x) = 125^x

Domain: all the real numbers (from negative infinity to infinity)

Range: all the positive real numbers.

As x decreases 125^x decreases approaching to 0. The function will never reach the value 0 but will approximate to it as x goes to negative infinity. The rate of decreasing gets lower and lower, with a limit of zero, when x becomes more and more negative.

As x increases 125^x increases indefinetely and at an increasing rate. That means that the increase is accelerated (the bigger the value of x the bigger the rate of increasing).
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What is the difference? StartFraction 2 x + 5 Over x squared minus 3 x EndFraction minus StartFraction 3 x + 5 Over x cubed minu
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\frac{(x+5)(x+2)}{x(x-3)(x+3)}

Step-by-step explanation:

The given expression is

\frac{2x+5}{x^{2} -3x} -\frac{3x+5}{x^{3} -9x} -\frac{x+1}{x^{2}-9}

First, we need to factor each denominator

\frac{2x+5}{x(x-3)} -\frac{3x+5}{x(x+3)(x-3)} -\frac{x+1}{(x-3)(x+3)}

So, the least common factor (LCF) is x(x-3)(x+3), because they are the factors that repeats.

Now, we diviide the LCF by each denominator, to then multiply it by each numerator.

\frac{(x+3)(2x+5)-3x-5-x(x+1)}{x(x-3)(x+3)} =\frac{2x^{2}+5x+6x+15-3x-5-x^{2}-x }{x(x-3)(x+3)}\\\frac{x^{2}+7x+10}{x(x-3)(x+3)}

Then, we factor the numerator, to do so, we need to find two numbers which product is 10 and which sum is 7.

\frac{(x+5)(x+2)}{x(x-3)(x+3)}

Therefore, the expression is equivalent to

\frac{(x+5)(x+2)}{x(x-3)(x+3)}

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