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Answer:
f^-1(x) = 4+∛((x-6)/5)
Step-by-step explanation:
To find the inverse function, solve ...
x = f(y)
then write the answer in functional form.
![\displaystyle x=f(y)\\\\x=5(y-4)^3+6\\\\x-6=5(y-4)^3\\\\\frac{x-6}{5}=(y-4)^3\\\\\sqrt[3]{\frac{x-6}{5}}=y-4\\\\y=4+\sqrt[3]{\frac{x-6}{5}}\\\\\boxed{f^{-1}(x)=4+\sqrt[3]{\frac{x-6}{5}}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%3Df%28y%29%5C%5C%5C%5Cx%3D5%28y-4%29%5E3%2B6%5C%5C%5C%5Cx-6%3D5%28y-4%29%5E3%5C%5C%5C%5C%5Cfrac%7Bx-6%7D%7B5%7D%3D%28y-4%29%5E3%5C%5C%5C%5C%5Csqrt%5B3%5D%7B%5Cfrac%7Bx-6%7D%7B5%7D%7D%3Dy-4%5C%5C%5C%5Cy%3D4%2B%5Csqrt%5B3%5D%7B%5Cfrac%7Bx-6%7D%7B5%7D%7D%5C%5C%5C%5C%5Cboxed%7Bf%5E%7B-1%7D%28x%29%3D4%2B%5Csqrt%5B3%5D%7B%5Cfrac%7Bx-6%7D%7B5%7D%7D%7D)
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The graph shows the function and its inverse to be reflections of each other in the line y=x, as they should be.
Hello!
if x = -3, just substitute it for x (replace the x with -3)
2(-3) + 6 = 0
2 * -3 = -6
-6 + 6 = 0
I hope this helps, and have a nice day!
Perhaps the easiest way to find the midpoint between two given points is to average their coordinates: add them up and divide by 2.
A) The midpoint C' of AB is
.. (A +B)/2 = ((0, 0) +(m, n))/2 = ((0 +m)/2, (0 +n)/2) = (m/2, n/2) = C'
The midpoint B' is
.. (A +C)/2 = ((0, 0) +(p, 0))/2 = (p/2, 0) = B'
The midpoint A' is
.. (B +C)/2 = ((m, n) +(p, 0))/2 = ((m+p)/2, n/2) = A'
B) The slope of the line between (x1, y1) and (x2, y2) is given by
.. slope = (y2 -y1)/(x2 -x1)
Using the values for A and A', we have
.. slope = (n/2 -0)/((m+p)/2 -0) = n/(m+p)
C) We know the line goes through A = (0, 0), so we can write the point-slope form of the equation for AA' as
.. y -0 = (n/(m+p))*(x -0)
.. y = n*x/(m+p)
D) To show the point lies on the line, we can substitute its coordinates for x and y and see if we get something that looks true.
.. (x, y) = ((m+p)/3, n/3)
Putting these into our equation, we have
.. n/3 = n*((m+p)/3)/(m+p)
The expression on the right has factors of (m+p) that cancel*, so we end up with
.. n/3 = n/3 . . . . . . . true for any n
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* The only constraint is that (m+p) ≠ 0. Since m and p are both in the first quadrant, their sum must be non-zero and this constraint is satisfied.
The purpose of the exercise is to show that all three medians of a triangle intersect in a single point.
