If there is more to the answer than what I give please ask for more!
With
we can find 
by substituting 4 into the main function.
So it would look like this:
Hope this helps!
With
So it would look like this:
Hope this helps!
If a = 10, then what is the value of 6a + 15?
2 answers:
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The answer is ![f^{-1}(x) = \sqrt[3]{ \frac{x-3}{6}}](https://tex.z-dn.net/?f=%20f%5E%7B-1%7D%28x%29%20%3D%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7Bx-3%7D%7B6%7D%7D)
Our function is f(x) = 6x³ + 3. We need to find f⁻¹(x)
Step 1: Substitute y for f(x):
y = 6x³ + 3
Step 2: Switch x and y because every (x, y) has its partner (y, x):
x = 6y³ + 3
Step 3: Solve for y:
![6y^{3} =x-3 \\ y^{3}= \frac{x-3}{6} \\ y = \sqrt[3]{ \frac{x-3}{6}}](https://tex.z-dn.net/?f=6y%5E%7B3%7D%20%3Dx-3%20%5C%5C%20%0Ay%5E%7B3%7D%3D%20%5Cfrac%7Bx-3%7D%7B6%7D%20%5C%5C%20%0Ay%20%3D%20%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7Bx-3%7D%7B6%7D%7D%20%20)
![f^{-1}(x) = \sqrt[3]{ \frac{x-3}{6}}](https://tex.z-dn.net/?f=%20f%5E%7B-1%7D%28x%29%20%3D%20%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7Bx-3%7D%7B6%7D%7D)
Our function is f(x) = 6x³ + 3. We need to find f⁻¹(x)
Step 1: Substitute y for f(x):
y = 6x³ + 3
Step 2: Switch x and y because every (x, y) has its partner (y, x):
x = 6y³ + 3
Step 3: Solve for y: