Question

The functions f and g are differentiable for all real numbers x. The table below gives values for the functions and their first derivatives at selected values of x.
x f(x) f'(x) g(x) g'(x)
1 4 -3 5 2
2 -3 -1 4 6
3 π 8 -1 4
4 -5 unknown 0 3
a. If the function h is given by h(x) = f (x) / g(x)' find h'(1).
b. If the function r is given by r(x) = f (x)g(x), find the equation of the tangent line to r(x) at x = 2.

Answer 1

Answer:

13/9

y =-22x+32

Step-by-step explanation:

Given that the functions f and g are differentiable for all real numbers x. The table below gives values for the functions and their first derivatives at selected values of x.

x f(x) f'(x) g(x) g'(x)

1 4 -3 5 2

2 -3 -1 4 6

3 π 8 -1 4

4 -5 unknown 0 3

a) $h(x) = \frac{f(x)}{g(x)} \\h'(x) = \frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}$

(using quotient rule)

Substitute 1 for x

$h'(1) = \frac{g(1)f'(1)-f(1)g'(1)}{(g(1))^2}\\=\frac{9-(-4)}{9} \\=\frac{13}{9}$

b) $r(x) = f(x) g(x)\\r'(x) = f(x) g'(x)+g(x)f'(x)$

when $x =2, r(x) = r(2) = f(2) g(2) = -12$

point of contact is (2,-12)

Slope of tangent =$r'(2) = f(2) g'(2)+g(2)f'(2)\\=-18+(-4) \\=-22$

Using point slope form, tangent is

$y+12 = -22(x-2)\\y = -22x +32$

Answer 2

The value of h'(1) is 13/9 and the equation of a tangent line to r(x) at (x = 2) is (y = -22x + 32) and this can be determined by using differentiation and by using the slope-intercept form.

Given :

• The functions f and g are differentiable for all real numbers x.
• The given table gives values for the functions and their first derivatives at selected values of x.

A)

$\rm h(x) =\dfrac{f(x)}{g(x)}$

$\rm h'(x) =\dfrac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}$

Now, at (x = 1) the above expression becomes:

$\rm h'(1) =\dfrac{g(1)f'(1)-f(1)g'(1)}{(g(1))^2}$

$\rm h'(1) =\dfrac{9-(-4)}{(3)^2}$

$\rm h'(1) =\dfrac{13}{9}$

B) r(x) = f(x)g(x)

r'(x) = f'(x)g(x) + f(x)g'(x)

Now, at (x = 2) the above expression becomes:

r'(2) = f'(2)g(2) + f(2)g'(2)

r'(2) = -18 + (-4)

r'(2) = -22

Using the one point form of the line, the tangent is given by:

$y+12=-22(x-2)$

y = -22x + 32

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https://brainly.com/question/21835898