Question
If $f^{\prime}(x)=\frac{1}{1+x^{2}}$ for all $x$ and $f(0)=0$, then(A) $f(2)<0.4$(B) $f(2)>2$(C) $0.4<f(2)<2$(D) $f(2)=2$
Step 1
This implies that $f(x)$ is continuous at all $x$. Show more…
Show all steps
Your feedback will help us improve your experience
Varsha Aggarwal and 75 other Calculus 1 / AB 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
If $f(x)=\left\{\begin{array}{ll}{x^{2}} & {\text { if } x<0} \\ {2} & {\text { if } x=0} \\ {2 x+1} & {\text { if } x>0}\end{array}\right.$ find: (a) $f(-2)$ (b) $f(0)$ (c) $f(2)$
Functions and Their Graphs
Library of Functions; Piecewise-defined Functions
If $f(x)=\left\{\begin{array}{ll}-x^{2} & \text { if } x<0 \\ 4 & \text { if } x=0 \text { find: } \\ 3 x-2 & \text { if } x>0\end{array}\right.$ (a) $f(-3)$ (b) $f(0)$ (c) $f(3)$
If $f(x)=\left\{\begin{array}{l}2+x, x \geq 0 \\ 2-x, x<0\end{array}\right.$, then $f(f(x))$ is given by (a) $\mathrm{f}(\mathrm{f}(\mathrm{x}))=\left\{\begin{array}{l}4+\mathrm{x}, \mathrm{x} \geq 0 \\ 4-\mathrm{x}, \mathrm{x}<0\end{array}\right.$ (b) $\mathrm{f}(\mathrm{f}(\mathrm{x}))=\left\{\begin{array}{c}4+\mathrm{x}, \mathrm{x} \geq 0 \\ \mathrm{x}, \mathrm{x}<0\end{array}\right.$ (c) $f(f(x))=\left\{\begin{array}{c}4-x, x \geq 0 \\ x, x<0\end{array}\right.$ (d) $\mathrm{f}(\mathrm{f}(\mathrm{x}))=\left\{\begin{array}{l}4+2 \mathrm{x}, \mathrm{x} \geq 0 \\ 4-2 \mathrm{x}, \mathrm{x}<0\end{array}\right.$
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD