Why is $y=b^x$ not continuous for $b>0$ and $b\ne 1$?

Question

The lines graphed by such do not have an interrupted domain.

So why is my textbook telling me this?

E: Typo in title. See photo.

Answer 1

$b$ is a fixed number in the function, not the variable.

What your textbook says is that, for every (fixed) real number $b$ such that $b>0$ and $b\neq 1$, the function $y=b^x$ is a continuous function on the real line. The domain of this function is the set of all real numbers. No interruptions.

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For exponential function (and the logarithm) the fixed number $b$ is called the basis. We do not consider the case $b=1$ simply because it is trivial (a constant function) and this trivial case behaves rather differently from the nontrivial cases.

Answer 2

$y=b^x$ is continuous, the definition of continuous function is Let $f: [a,b] \to \mathbb{R}$ is continuous in a point $c \in I$ if $\forall \varepsilon >0$ there exists $\delta>0$ such that When $|x-c|<0$ then $|f(x)-f(c)|< \varepsilon$ Or actually true $f $ is continuous function in $c$ if $\lim_{x \to c} f(x)=f(c)$ And ass other comments say the function $y=b^x$ is continuous for $b\neq 0$ And for $b=1$ the function is constant

Answer 3

Note that, for $b=1$ the function $y = \log_1 x$ is actually not defined for any $x \not =1$, as $1^y =1$ for all real $y$. For real $x \not =1$ there is no $y$ satisfying $1^y = x$. However for every other positive $b >0$ and every other positive $x$ the function $y = \log_b x$ is well-defined; equivalently there is a real number $y$ such that $b^y = x$.

However, $b^x$ is indeed defined for all $b >0$ including $b=1$. It is the horizontal line $y=1$.

The likely reason why $b =1$ was excluded even for the function $y = b^x$ is that for precisely $b>0$; $b \not =1$ the functions $y=b^x$ and $y=\log_b x$ are invertible; in fact they pair up as each other's inverse; if $y=b^x$ then $x = \log_b y$.