differentiating inverse functions\n18 the functions f and g are differentiable for all real numbers and g is…

differentiating inverse functions\n18 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.\n| x | f(x) | f(x) | g(x) | g(x) |\n|----|------|--------|------|--------|\n| 1 | 6 | 4 | 2 | 5 |\n| 2 | 9 | 2 | 3 | 1 |\n| 3 | 10 | -4 | 4 | 2 |\n| 4 | -1 | 3 | 6 | 7 |\n(a) explain why there must be a value r for 1 < r < 3 such that h(r)= - 5.\n(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.\n19 a function h satisfies h(3)=5 and h(3)=7. which of the following statements about the inverse of h must be true?\n(a) (h^(-1))(5)=3\n(b) (h^(-1))(7)=3\n(c) (h^(-1))(5)=7\n(d) (h^(-1))(5)=1/7\n(e) (h^(-1))(7)=1/5

differentiating inverse functions\n18 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.\n| x | f(x) | f(x) | g(x) | g(x) |\n|----|------|--------|------|--------|\n| 1 | 6 | 4 | 2 | 5 |\n| 2 | 9 | 2 | 3 | 1 |\n| 3 | 10 | -4 | 4 | 2 |\n| 4 | -1 | 3 | 6 | 7 |\n(a) explain why there must be a value r for 1 < r < 3 such that h(r)= - 5.\n(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.\n19 a function h satisfies h(3)=5 and h(3)=7. which of the following statements about the inverse of h must be true?\n(a) (h^(-1))(5)=3\n(b) (h^(-1))(7)=3\n(c) (h^(-1))(5)=7\n(d) (h^(-1))(5)=1/7\n(e) (h^(-1))(7)=1/5

Answer

(a)

Explanation:

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)

Explanation:

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)

Explanation:

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})