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Is this statement correct

Okay, is the above statement correct? Because if I put $x=\frac{1}{2}$ then $f(x)$ will have two values. So will that remain a function anymore?

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  • $\begingroup$ Working with exponential functions (as defined above), we only consider the principal root. For example, $a^{\frac{11}{4}}$ could be interpreted as $\sqrt[4]{a^{11}}$. This is perfect for defining a function. But you can't use this notion everywhere. Consider $(-3)^2=9$, which means $(-3)^{2\cdot\frac{1}{2}}=9^{\frac{1}{2}}$, so $(-3)^1=-3=3$? This is wrong of course. $\endgroup$ – Corellian Oct 19 '18 at 5:39
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Indeed, the unrestricted inverse of any non-injective function is multivalued, but this is fixed by simply restricting the co-domain to some interval. Take $f(x)=y=x^2$ for example. It's inverse satisfies $f^{-1}(y) = x$. Therefore, if we just solve our original function for $x$, we should have our answer.

$$x^2 = y \\ x^2 - y = 0 \\ (x+y^{\frac{1}{2}})(x-y^{\frac{1}{2}})=0 \\ f^{-1}(y) = x = \pm y^{\frac{1}{2}}$$

Clearly, this is multivalued, and cannot be a function, so by convention, we define the principal square root function to be positive solution to be the inverse of $f(x) = y = x^2$.

In conclusion, the statement is correct because $f(x)=a^x$ only has one value at $f(\frac{1}{2})$.

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