Question

PLEASE HELP!! DERIVATIVES AND GRAPHS

Answer 1

Observe the graph f(x) is a absolute function.

Vertex of f(x) (2, 4) and it's look like downward v.

So the function f(x) = - 2 |x - 2| + 4.

Observe the graph g(x) is a linear function.

x intercept of g(x) 4 and y intercept of g(x) is 4.

The line equation with intercepts x/a + y/b = 1

x/4 + y/4 = 1

(x + y)/4 = 1

x + y = 4

y = - x + 4

So the function g(x) = - x + 4.

h(x) = f(x) g(x)

h(x) = (- 2 |x - 2| + 4)(- x + 4)

h(x) = (- 2 |x - 2|)(- x) + 4(- x) + (- 2 |x - 2|)(4) + 4(4)

h(x) = (2x |x - 2|) - 4x - (8|x - 2|) + 16

Apply derivative on each side with respect of x.

h'(x) = d/dx [(2x |x - 2|) - 4x - (8|x - 2|) + 16]

h'(x) = d/dx [(2x |x - 2|)] + d/dx(- 4x) - d/dx(8|x - 2|) + d/dx (16)

h'(x) = d/dx [(2x |x - 2|)] - 4 - d/dx(8|x - 2|)

Derivative of absolute value = |u|

du/dx = [u/|u|]u'

d/dx(8|x - 2|) = 8((x-2)/|x - 2|) (1)

= (8x - 16)/|x - 2|

d/dx [(2x |x - 2|)] = (2x)((x - 2)/|x - 2|) + | x - 2|(2)

= (2x² - 4x)/|x - 2| + 2|x - 2|

h'(x) = {(2x² - 4x)/(|x - 2|) + 2|x - 2|} - 4 - {(8x - 16)/|x - 2|}

a) h'(1) = {(2(1)² - 4(1))/(|1 - 2|) + 2|1 - 2|} - 4 - {(8(1) - 16)/|1 - 2|}

h'(1) = (2 - 4) + 2 - 4 - (-8)

h'(1) = - 2 + 2 - 4 + 8

h'(1) = 4

 

b) h'(2) = {(2(2)² - 4(2))/(|2 - 2|) + 2|2 - 2|} - 4 - {(8(1) - 16)/|2 - 2|}

h'(2) does not exist.

h' (2) = DNE

 

c) h'(3) = {(2(3)² - 4(3))/(|3 - 2|) + 2|3 - 2|} - 4 - {(8(3) - 16)/|3 - 2|}

h'(3) = 18 - 12 + 2 - 4 - (24 - 16)

h'(3) = 18 - 12 + 2 - 4 - 8

h'(3) = - 4.