Number and Algebra
Functions
Geometry & Trigonometry
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Calculus
Suppose fff is invertible and f(2)=5f(2)=5f(2)=5, f(3)=6f(3)=6f(3)=6, f(6)=8f(6)=8f(6)=8, f(16)=18f(\tfrac{1}{6})=18f(61)=18. Calculate f−1(8)f^{-1}(8)f−1(8).
Let fff be an invertible function such that f(2)=5f(2)=5f(2)=5, f(3)=6f(3)=6f(3)=6, f(6)=8f(6)=8f(6)=8 and f(16)=18f(\tfrac{1}{6})=18f(61)=18. Find f−1(6)f^{-1}(6)f−1(6).
Let fff be an invertible function with f(2)=5f(2)=5f(2)=5, f(3)=6f(3)=6f(3)=6, f(6)=8f(6)=8f(6)=8 and f(16)=18f(\tfrac{1}{6})=18f(61)=18. Determine f−1(5)f^{-1}(5)f−1(5).
Given an invertible function fff with f(2)=5f(2)=5f(2)=5, f(3)=6f(3)=6f(3)=6, f(6)=8f(6)=8f(6)=8 and f(16)=18f(\tfrac{1}{6})=18f(61)=18, find f−1(18)f^{-1}(18)f−1(18).
Let fff be invertible with f(2)=5f(2)=5f(2)=5, f(3)=6f(3)=6f(3)=6, f(6)=8f(6)=8f(6)=8, f(16)=18f(\tfrac{1}{6})=18f(61)=18. Evaluate f(f−1(6))f\bigl(f^{-1}(6)\bigr)f(f−1(6)).
With fff invertible and f(2)=5f(2)=5f(2)=5, f(3)=6f(3)=6f(3)=6, f(6)=8f(6)=8f(6)=8, f(16)=18f(\tfrac{1}{6})=18f(61)=18, compute f−1(f(16))f^{-1}\bigl(f(\tfrac{1}{6})\bigr)f−1(f(61)).
Let fff be invertible with f(2)=5f(2)=5f(2)=5, f(3)=6f(3)=6f(3)=6, f(6)=8f(6)=8f(6)=8, f(16)=18f(\tfrac{1}{6})=18f(61)=18. If f−1(x)=3f^{-1}(x)=3f−1(x)=3, find xxx.
Let h(x)=f(f−1(x)−1)h(x)=f\bigl(f^{-1}(x)-1\bigr)h(x)=f(f−1(x)−1), where fff is invertible and f(2)=5f(2)=5f(2)=5, f(3)=6f(3)=6f(3)=6, f(6)=8f(6)=8f(6)=8, f(16)=18f(\tfrac{1}{6})=18f(61)=18. Determine h(6)h(6)h(6).
Define g(x)=f−1(x)+2g(x)=f^{-1}(x)+2g(x)=f−1(x)+2 for an invertible fff with f(2)=5f(2)=5f(2)=5, f(3)=6f(3)=6f(3)=6, f(6)=8f(6)=8f(6)=8, f(16)=18f(\tfrac{1}{6})=18f(61)=18. Find g(6)g(6)g(6).
Suppose fff is invertible and f(2)=5f(2)=5f(2)=5, f(3)=6f(3)=6f(3)=6, f(6)=8f(6)=8f(6)=8, f(16)=18f(\tfrac{1}{6})=18f(61)=18. Evaluate f−1(f(f−1(8)))f^{-1}\bigl(f\bigl(f^{-1}(8)\bigr)\bigr)f−1(f(f−1(8))).
Define p(x)=f−1(x)×f−1(f(x))p(x)=f^{-1}(x)\times f^{-1}\bigl(f(x)\bigr)p(x)=f−1(x)×f−1(f(x)) for invertible fff with f(2)=5f(2)=5f(2)=5, f(3)=6f(3)=6f(3)=6, f(6)=8f(6)=8f(6)=8, f(16)=18f(\tfrac{1}{6})=18f(61)=18. Compute p(6)p(6)p(6).
Let k(x)=f−1(f(x)+12)k(x)=f^{-1}\bigl(f(x)+12\bigr)k(x)=f−1(f(x)+12) for an invertible fff with f(2)=5f(2)=5f(2)=5, f(3)=6f(3)=6f(3)=6, f(6)=8f(6)=8f(6)=8, f(16)=18f(\tfrac{1}{6})=18f(61)=18. Find k(16)k(\tfrac{1}{6})k(61).
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