How do you sketch the graph that satisfies f'(x)=1 when x-2, f'(x)=-1 when x

Question

How do you sketch the graph that satisfies f'(x)=1 when x>-2, f'(x)=-1 when x<-2, f(-2)=-4?

Luke Alvarez · Accepted Answer

# f(x) = { (-x+a, x < -2), (-4, x = -2), (x+b, x > -2) :}# where #a,b# are constants

For #x < -2# we have #f'(x)=-1 => f(x)=-x+a#, ie a straight line with gradient #-1#

For #x > -2# we have #f'(x)=1 => f(x)=x+b#, ie a straight line with gradient #1#

We want #f(-2)=-4# and we are not told anything about the gradient when #x=-2# (and in fact #f'(-2)# will be undefined) and we are not told that that the function should be continuous.

Combining the results we get:

# f(x) = { (-x+a, x < -2), (-4, x = -2), (x+b, x > -2) :}# where #a,b# are constants
If we wanted a continuous solutions then we would need:

For #x < -2 => 2+a = -4 \ \ \ \ => a=-6 #
For #x > -2 => -2+b = -4 => b=-2 #

And so:

# f(x) = { (-x-6, x < -2), (-4, x = -2), (x-2, x > -2) :}#

Which we can graph as follows:

But equally we could choose any values for the arbitrary constants #a# and #b#, for example we could choose #a=b=0# to get

# f(x) = { (-x, x < -2), (-4, x = -2), (x, x > -2) :}#

Which we can graph as follows:

Emilia Aguilar · Answer

undefined To sketch the graph that satisfies $ f'(x) = 1 $ when $ x > -2 $, $ f'(x) = -1 $ when $ x < -2 $, and $ f(-2) = -4 $, follow these steps: 1. For $ x > -2 $, the function $ f'(x) = 1 $ implies that the function $ f(x) $ is increasing. So, draw a straight line with a slope of 1 for $ x > -2 $. 2. For $ x < -2 $, the function $ f'(x) = -1 $ implies that the function $ f(x) $ is decreasing. So, draw a straight line with a slope of -1 for $ x < -2 $. 3. Connect the two lines smoothly at $ x = -2 $ since $ f(-2) = -4 $. 4. Label the point $ (-2, -4) $ on the graph. This graph represents the function $ f(x) $ that satisfies the given conditions.

How do you sketch the graph that satisfies f'(x)=1 when x&gt;-2, f'(x)=-1 when x&lt;-2, f(-2)=-4?

How do you sketch the graph that satisfies f'(x)=1 when x>-2, f'(x)=-1 when x<-2, f(-2)=-4?