Question
Suppose $f(x)=\frac{x^{2}-6 x-16}{\left(x^{2}-7 x-8\right) \sqrt{x^{2}-4}}$(a) For what numbers $x$ is $f$ defined?(b) For what numbers $x$ is $f$ discontinuous?(c) Which discontinuities found in (b) are removable?
Step 1
The function is defined for all $x$ such that the denominator of the function is not equal to zero and the expression under the square root is non-negative. Show more…
Show all steps
Your feedback will help us improve your experience
Carson Merrill and 62 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
Find the points of discontinuity, and determine whether the discontinuities are removable. (a) $f(x)=\frac{x^{2}-4}{x^{3}-8}$ (b) $f(x)=\left\{\begin{array}{ll}2 x-3, & x \leq 2 \\ x^{2}, & x>2\end{array}\right.$ (c) $f(x)=\left\{\begin{array}{ll}3 x^{2}+5, & x \neq 1 \\ 6, & x=1\end{array}\right.$
Limits and Continuity
Continuity
Find the values of $x$ (if any) at which $f$ is not continuous, and determine whether each such value is a removable discontinuity. $$ \begin{array}{ll}{\text { (a) } f(x)=\frac{x^{2}-4}{x^{3}-8}} & {\text { (b) } f(x)=\left\{\begin{array}{ll}{2 x-3,} & {x \leq 2} \\ {x^{2},} & {x>2}\end{array}\right.} \\ {\text { (c) } f(x)=\left\{\begin{array}{ll}{3 x^{2}+5,} & {x \neq 1} \\ {6,} & {x=1}\end{array}\right.}\end{array} $$
LIMITS AND CONTINUITY
(a) find each point of discontinuity. (b) Which of the discontinuities are removable? not removable? Give reasons for your answers. $$ f(x)=\left\{\begin{array}{ll} 1-x^{2}, & x \neq-1 \\ 2, & x=-1 \end{array}\right. $$
limits and Continuity
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD