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Recall the inverse function theorem: if f(x) has an inverse, and if f(a) = b and a = f⁻¹(b), then
f⁻¹(f(x)) = x ⇒ (f⁻¹)'(f(x)) • f'(x) = 1 ⇒ (f⁻¹)'(f(x)) = 1/f'(x)
⇒ (f⁻¹)'(b) = 1/f'(a)
Let b = 10. Then pick the function f(x) such that f(a) = 10 and f'(a) = -8 for some number a.
The decreasing function which f⁻¹'(10) = -1/8 is f(x) = -2x³ - 2x + 14
To do this, we make use of the following inverse function theorem.
Given that f(a) = b then a = f⁻¹(b)
The above means that:
If f⁻¹(10) = 1/8
Then f(1/8) = 10
From the list of options, we have:
f(x) = -2x³ - 2x + 14
Set f(x) = 10
-2x³ - 2x + 14 = 10
Subtract 10 from both sides
-2x³ - 2x + 4 = 0
Divide through by -2
x³ + x - 2 = 0
Expand
(x - 1)(x² + x + 2) = 0
Split
x - 1 = 0 or x² + x + 2 = 0
The equation x² + x + 2 = 0 has no real solution
So, we have:
x - 1 = 0
Solve for x
x = 1
Differentiate function f(x)
f'(x) = -(6x² + 2)
Take the inverse of both sides
[tex]\frac{1}{f'(x)} = \frac{1}{-(6x\² + 2)}[/tex]
Substitute 10 for x
[tex]\frac{1}{f'(1)} = \frac{1}{-(6 * 1\² + 2)}[/tex]
[tex]\frac{1}{f'(1)} = -\frac{1}{8}[/tex]
This means that:
f⁻¹'(10) = -1/8
Hence, the function which f⁻¹'(10) = -1/8 is f(x) = -2x³ - 2x + 14
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