Question
(a) Show that $f$ is one-to-one.(b) Use Theorem 7 to find $\left(f^{-1}\right)^{\prime}(a)$.(c) Calculate $f^{-1}(x)$ and state the domain and range of $f^{-1}$.(d) Calculate $\left(f^{-1}\right)^{\prime}(a)$ from the formula in part (c) and check that it agrees with the result of part (b).(e) Sketch the graphs of $f$ and $f^{-1}$ on the same axes.$f(x)=x^3, \quad a=8$
Step 1
A function is one-to-one if for any $x_1, x_2$ in the domain, $f(x_1) = f(x_2)$ implies $x_1 = x_2$. Suppose $f(x_1) = f(x_2)$, then $x_1^3 = x_2^3$. Taking the cube root of both sides: $x_1 = x_2$. Therefore, $f(x) = x^3$ is one-to-one. Show more…
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(a) Show that $f$ is one-to-one. (b) Use Theorem 7 to find $\left(f^{-1}\right)^{\prime}(a)$ . (c) Calculate $f^{-1}(x)$ and state the domain and range of $f^{-1}$ (d) Calculate $f^{-1}(x)$ and state the formula in part (c) and check that it agrees with the result of part (b). (e) Sketch the graphs of $f$ and $f^{-1}$ on the same axes. $$ f(x)=x^{3}, \quad a=8 $$
INVERSE FUNCTIONS
Inverse Functions and Logarithms
Need to do A-E for37
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