Question

I don’t know how to do this…

Answer 1

Step-by-step explanation:

f^-1 means the inverse function that finds the x responsible for the creation of a given y (functional result) value.

y = 3x - 1

3x = y + 1

x = 1/3(y + 1)

and now we rename x to y and y to x, so that we get a normal function definition :

y = 1/3(x + 1) = (1/3)x + 1/3

now we create the first derivative d/dx of this function :

y' = 1/3

since both f(x) and f^‐1(x) are line functions with a constant slope, it is not surprising that the first derivative (the function of the slope of the base function) is a constant function.

so, for any value of x (incl. the requested x = 11) d/dx f^-1(x) = 1/3.

therefore also d/dx f^-1(11) = 1/3.

y = 2×sqrt(x)

sqrt(x) = y/2

x = y²/4

again we rename the variables, so that we get a normal function definition for the inverse function :

y = x²/4

the first derivative is :

y' = 2x/4 = x/2

so, d/dx f^-1(6) = 6/2 = 3

Answer 2

Use the inverse function theorem: if f(x) is continuously differentiable at x = a, and given f(a) = b, then f is invertible around x = a, and

$\left(f^{-1}\right)'(b) = \dfrac1{f'(a)}$

Both the given functions f(x) meet the conditions above. Note that you don't need to actually compute the inverse function itself.

If f(x) = 3x - 1, then f(x) = 11 when

3x - 1 = 11   ⇒   3x = 12   ⇒   x = 4

Now if f(4) = 11, it follows from the inverse function theorem that

$\left(f^{-1}\right)'(11) = \dfrac1{f'(4)} = \boxed{\dfrac13}$

since f'(x) = 3.

If f(x) = 2√x, then f(x) = 6 when

2√x = 6   ⇒   √x = 3   ⇒   x = ±9

But note that f(x) is defined only for x ≥ 0, so we must take x = 9. Now if f(9) = 6, the inverse function theorem says

$\left(f^{-1}\right)'(6) = \dfrac1{f'(9)} = \dfrac1{\frac1{\sqrt9}} = \boxed{3}$

since f'(x) = 1/√x.