Question
A function $f: R^{+} \rightarrow[0,1)$ is defined as $f(x)=\frac{x^{2}}{x^{2}+1}$Then find $f^{-1}(x)$
Step 1
We can rewrite this as $f(x)=1-\frac{1}{x^{2}+1}$. Show more…
Show all steps
Your feedback will help us improve your experience
Rukhmani Jain and 86 other Precalculus educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
A function $f: R^{+} \rightarrow[0,1]$ is defined as $f(x)=\frac{x^{2}}{x^{2}+1}$. Find $f^{-1}(x)$.
Real Function
Level I
A function $\mathrm{f}$ is defined as $f(x)=\frac{1}{x^{2}+1}$ where f: $R^{+} \cup$ $\{0\} \rightarrow(0,1]$, find $f^{\prime}(x)$
Inverse Trigonometric Functions
A function $f(x)$ is defined as $f(x)=\left(a-x^{n}\right)^{1 / n}, x>0, n \in I^{+}$ Then find $f(f(x))+f\left(f\left(\frac{1}{x}\right)\right)$.
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD