Question

differentiating inverse functions
18 the functions f and g are differentiable for all real numbers and g is strictly increasing. the table below gives values of the functions and their first derivatives at selected values of x. the function h is given by h(x)=f(g(x)) - 6.

x	f(x)	f(x)	g(x)	g(x)
2	9	2	3	1
3	10	-4	4	2
4	-1	3	6	7

(a) explain why there must be a value r for 1 < r < 3 such that h(r)= - 5.
(b) if g^(-1) is the inverse function of g, write an equation for the line tangent to the graph of y = g^(-1)(x) at x = 2.
19 a function h satisfies h(3)=5 and h(3)=7. which of the following statements about the inverse of h must be true?
(a) (h^(-1))(5)=3
(b) (h^(-1))(7)=3
(c) (h^(-1))(5)=7
(d) (h^(-1))(5)=1/7
(e) (h^(-1))(7)=1/5

Explanation:

(a)

Step1: Find \(h(1)\) and \(h(3)\)

First, use the formula \(h(x)=f(g(x)) - 6\). When \(x = 1\), \(g(1)=2\), then \(h(1)=f(g(1))-6=f(2)-6\). From the table, \(f(2) = 9\), so \(h(1)=9 - 6=3\). When \(x = 3\), \(g(3)=4\), then \(h(3)=f(g(3))-6=f(4)-6\). From the table, \(f(4)=-1\), so \(h(3)=-1 - 6=-7\).

Step2: Apply Intermediate - Value Theorem

The functions \(f\) and \(g\) are differentiable for all real numbers, so \(h(x)\) is continuous on the interval \([1,3]\) (since composition and subtraction of continuous - differentiable functions is continuous). We have \(h(1) = 3\) and \(h(3)=-7\), and \(-5\) is between \(3\) and \(-7\). By the Intermediate - Value Theorem, which states that if \(y = h(x)\) is continuous on the closed interval \([a,b]\) and \(k\) is a number between \(h(a)\) and \(h(b)\), then there exists at least one number \(r\) in the open interval \((a,b)\) such that \(h(r)=k\). Here, \(a = 1\), \(b = 3\), and \(k=-5\), so there must be a value \(r\) for \(1\lt r\lt3\) such that \(h(r)=-5\).

(b)

Step1: Recall the formula for the derivative of an inverse function

The formula for the derivative of the inverse function \(y = g^{-1}(x)\) is \((g^{-1})'(x)=\frac{1}{g'(g^{-1}(x))}\). We want to find the equation of the tangent line to \(y = g^{-1}(x)\) at \(x = 2\). First, we need to find \(g^{-1}(2)\) and \((g^{-1})'(2)\). From the table, when \(x = 1\), \(g(1)=2\), so \(g^{-1}(2)=1\).

Step2: Calculate \((g^{-1})'(2)\)

Using the formula \((g^{-1})'(2)=\frac{1}{g'(g^{-1}(2))}\), and since \(g^{-1}(2)=1\) and \(g'(1)=5\), we have \((g^{-1})'(2)=\frac{1}{5}\).

Step3: Use the point - slope form of a line

The point - slope form of a line is \(y - y_1=m(x - x_1)\), where \((x_1,y_1)\) is a point on the line and \(m\) is the slope of the line. For the tangent line to \(y = g^{-1}(x)\) at \(x = 2\), \(x_1 = 2\), \(y_1=g^{-1}(2)=1\), and \(m=(g^{-1})'(2)=\frac{1}{5}\). The equation of the tangent line is \(y - 1=\frac{1}{5}(x - 2)\), which simplifies to \(y=\frac{1}{5}x+\frac{3}{5}\).

(c)

Step1: Recall the formula for the derivative of an inverse function

The formula for the derivative of the inverse function of \(y = h(x)\) is \((h^{-1})'(y)=\frac{1}{h'(h^{-1}(y))}\). We know that \(h(3)=5\), so \(h^{-1}(5)=3\).

Step2: Calculate \((h^{-1})'(5)\)

Using the formula \((h^{-1})'(5)=\frac{1}{h'(h^{-1}(5))}\), and since \(h^{-1}(5)=3\) and \(h'(3)=7\), we have \((h^{-1})'(5)=\frac{1}{7}\).

Answer:

(a) The function \(h(x)\) is continuous on \([1,3]\) (as a composition and subtraction of differentiable functions), \(h(1) = 3\), \(h(3)=-7\), and since \(-5\) is between \(3\) and \(-7\), by the Intermediate - Value Theorem, there exists \(r\in(1,3)\) such that \(h(r)=-5\).
(b) \(y=\frac{1}{5}x+\frac{3}{5}\)
(c) D. \((h^{-1})'(5)=\frac{1}{7}\)