A student finds the inverse of the function [tex]\( y=\frac{1}{16}(x-5)^2+9 \)[/tex] this way:

Equation:
[tex]\[ y=\frac{1}{16}(x-5)^2+9 \][/tex]

Step 1:
[tex]\[ x=\frac{1}{16}(y-5)^2+9 \][/tex]

Step 2:
[tex]\[ x-9=\frac{1}{16}(y-5)^2 \][/tex]

Step 3:
[tex]\[ 16(x-9)=(y-5)^2 \][/tex]

Step 4:
[tex]\[ 16 \sqrt{(x-9)}=\sqrt{(y-5)^2} \][/tex]

Step 5:
[tex]\[ 16 \sqrt{(x-9)}=y-5 \][/tex]

Step 6:
[tex]\[ 5+16 \sqrt{(x-9)}=y \][/tex]

The solution is incorrect. Identify the step that does not follow logically from the previous step.



Answer :

Let's carefully go through the steps to identify any logical errors in the student's solution for finding the inverse of the function [tex]\( y = \frac{1}{16}(x-5)^2 + 9 \)[/tex].

1. Original equation: [tex]\( y = \frac{1}{16}(x-5)^2 + 9 \)[/tex]

2. Step 1: Switch [tex]\( x \)[/tex] and [tex]\( y \)[/tex] to find the inverse:
[tex]\[ x = \frac{1}{16}(y-5)^2 + 9 \][/tex]

3. Step 2: Isolate the quadratic term to the left side:
[tex]\[ x - 9 = \frac{1}{16}(y-5)^2 \][/tex]

4. Step 3: Remove the fraction by multiplying both sides by 16:
[tex]\[ 16(x - 9) = (y - 5)^2 \][/tex]

5. Step 4: Apply the square root to both sides to solve for [tex]\( y - 5 \)[/tex]:
[tex]\[ \sqrt{16(x - 9)} = \sqrt{(y - 5)^2} \][/tex]

Here, the student has made a logical error. Taking the square root of both sides should take into account both the positive and negative roots of the quadratic, resulting in:
[tex]\[ \sqrt{16(x - 9)} = \pm (y - 5) \][/tex]

6. Step 5: The correction from Step 4 leads to:
[tex]\[ 4\sqrt{(x - 9)} = \pm (y - 5) \][/tex]

7. Step 6: Solve for [tex]\( y \)[/tex]:
[tex]\[ \pm 4\sqrt{(x - 9)} = y - 5 \][/tex]
This gives two possible solutions:
[tex]\[ y = 5 + 4\sqrt{(x - 9)} \][/tex]
and
[tex]\[ y = 5 - 4\sqrt{(x - 9)} \][/tex]

Therefore, the step that does not follow logically from the previous step is Step 4. The correct step 4 should include both the positive and negative square roots, leading to the realization that there are two potential expressions for [tex]\( y \)[/tex].