Math  /  Algebra

QuestionFind a formula for f1(x)f^{-1}(x). Identify the domain and range of f1f^{-1}. Verify that ff and f1f^{-1} are inverses. f(x)=x+5,x5f(x)=\sqrt{x+5}, x \geq-5 f1(x)=x25f^{-1}(x)=x^{2}-5 (Simplify your answer.) Identify the domain of f1f^{-1}. Select the correct choice below and, if necessary, fill in the answer box(es) to complete your A. D={xx}D=\{x \mid x \leq\} B. D={xx}D=\{x \mid \leq x \leq\} C. D={xx}D=\{x \mid x \neq\} D. D={xx0}D=\{x \mid x \geq 0\} E. The domain is all real numbers.
Identify the range of f1f^{-1}. Select the correct choice below and, if necessary, fill in the answer box(es) to complete your ch A. R={yy}R=\{y \mid \leq y \leq\} B. R={yy}R=\{y \mid y \leq \quad\} C. R={yy5}R=\{y \mid y \geq-5\} D. R={yy}R=\{y \mid y \neq\} E. The range is all real numbers.
To verify that ff and f1f^{-1} are inverses, determine (f1f)(x)\left(f^{-1} \circ f\right)(x) and (ff1)(x)\left(f \circ f^{-1}\right)(x). (ff1)(x)=(f1f)(x)=\left(f \circ f^{-1}\right)(\mathrm{x})=\square \quad\left(\mathrm{f}^{-1} \circ \mathrm{f}\right)(\mathrm{x})=\square (Simplify your answers.)

Studdy Solution
f1(x)=x25f^{-1}(x) = x^2 - 5.
The domain of f1f^{-1} is D={xx0}D = \{x \mid x \geq 0\}.
The range of f1f^{-1} is R={yy5}R = \{y \mid y \geq -5\}. (ff1)(x)=x(f \circ f^{-1})(x) = x and (f1f)(x)=x(f^{-1} \circ f)(x) = x.

View Full Solution - Free
Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord