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When is $g (x) = 0$ $g(x)=0$ for the function $g (x) = 5 \cdot 2^{3 x} + 4$ $g(x)=5*2^(3x)+4$ ?

2 Answers

Nov 8, 2016

If $g (x) = 5 \cdot 2^{3 x} + 4$ $g(x)=5 * 2^(3x)+4$
then $g (x)$ $g(x)$ is never $= 0$ $=0$

Explanation:

For any positive value $k$ $k$ and any Real value $p$ $p$
$XXX k^{p} > 0$ $color(white)("XXX")k^p > 0$

Therefore
$XXX 2^{3 x} > 0$ $color(white)("XXX")2^(3x) > 0$ for $AAx in RR$

and
$color(white)("XXX")rarr 5*2^(3x) > 0$ for $AAx in RR$

and
$color(white)("XXX")rarr 5*2(3x)+4 > 0$ for $AAx in RR$

Nov 8, 2016

For this function, $g(x) != 0$ .

Explanation:

This is an exponential function, and, generally, exponential functions have no $y$ -value equal to $0$ . This is because no exponent of any number will give you $0$ (or anything smaller than it).

The only way to have an exponential function which intercepts the $x$ -axis is the translate the graph downwards.

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