In Exercises 73 and 74, repeat the steps above to solve for th...

Question

In Exercises 73 and 74, repeat the steps above to solve for the functions y=f(x) and x=f^(-1)(y) defined implicitly by the given equations over the interval.

73. y^(1/3) - 1 = (x+2)³, -5 ≤ x ≤ 5, x_0 = -3/2

Accepted Answer

Welcome back everyone. Given the y to the power of 1/4 minus 2 equals x + 52 with x between 7 and 3 inclusive, find explicit formulas for fx and f inverse of x. So first of all, let's identify f of x. What we want to do is simply solve this equation for y. We have y to the power of 1/4 minus 2 equals x + 52. So what we can do is simply add 2 to both sides to get y to the power of 1/4 equals x + 52 + 2, right? And now we're going to raise both sides to the power of 4. To get y equals. X + 52 + 2 raise to the power of 4 and this is how we got f of x and this is our first answer for this problem. And now let's identify f inverse. Let's remember how we identify inverse functions. The first step is to swap x and y. So instead of y, we're going to get x. That will be x to the power of 1/4 minus 2 equals. On the right hand side, x becomesy, so we're going to have y + 5 squad. Now, what we're going to do is basically take square root of both sides. And we can show that Y + 5. Is going to be equal to positive or negative square root of. X to the power of 1/4 minus 2. Now the reason why we have two possible roots is because taking the square root of the square gives us the absolute value of that function. So we are going to have the absolute value of y + 5, which gives us two possible solutions, positive or negative. Now for inverse functions, we can only have one possibility, right? And to determine which one is correct, we have to remember that the range of our inverse is basically the domain of the original function. So if we solve for y, we can show that y is equal to positive or negative square root of. X to the power of 1/4 minus 2 minus 5, right? And now if we look at the domain of the original function which now becomes the range, we can show that y is between -7. And 3 inclusive, right? So, the minimum value is -7, and the maximum value is 3. So to get 3, we basically want to take the positive fruit. Well, why is this the case? Well, if we take the negative fruit and we know that the square root is always non-negative, we're going to have only negative numbers for y. Andy must also be positive. So to get the positive Y, we must take the positive root, right? So that we get 3 when the square root is equal to 8, so it must be positive. This is why we're taking the positive root, and now we know that our inverse function. Is equal to, let's label it as f inverse of x. It is going to be equal to the square root of we can write x to the power of 1/4 square root of x minus 2, and then we are subtracting 5. So those would be our final answers for this problem. Thank you for watching.

In Exercises 73 and 74, repeat the steps above to solve for the functions y=f(x) and x=f^(-1)(y) defined implicitly by the given equations over the interval.73. y^(1/3) - 1 = (x+2)³, -5 ≤ x ≤ 5, x_0 = -3/2

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In Exercises 73 and 74, repeat the steps above to solve for the functions y=f(x) and x=f^(-1)(y) defined implicitly by the given equations over the interval.
73. y^(1/3) - 1 = (x+2)³, -5 ≤ x ≤ 5, x_0 = -3/2