Question

suppose the functions f(x) is described by:
f(x)= x+B if x<1
Ax+3 if x> or equal to 1
a) find A and B so that f(x) is continuous for all numbers and f(-1)=0. briefly explain your answer

b) sketch y=f(x) on the axis given for the values A and B found in a) when x is in the interval [-2,2]

Answer 1

\[f(x)=\begin{cases}x+B~~~~~~~~~~~~~if~~x<1\\Ax+3~~~~~~~~~~~if~~x\geq1\end{cases}\]
Now, in the first case and part a) you have f(-1) =0 , that means x =-1 <0, right? so that f(x)= x+B and f(-1) =-1+B =0 --> B=1

Answer 2

right, i got that part. its mainly finding A that im having a hard time with

Answer 3

To have the f(x) continuous, you need some value of A such that the graph continuous.
One more thing: the value of x much be \(\geq 1\) and 1 is the "intersect" point which links 2 parts of f(x). Now, let consider f(1) = A*1+3
and you consider if x =1 f(1) = x+B = 1+1 =2 ( you need this value to consider A)
so A*1+3 =2 --> A =-1
so, if you have A =-1 and B =1 the graph is continuous.
follow is the graph.
\(\color {red}{\text{I am cheating by graphing it first}}\)

Answer 4

By using demos graph calculator, I replace the value of A such that the 2 lines intersects at 1.

Answer 5

on the graph the two lines intersect at 2 though

Answer 6

so i understand everything else except for how to graph it

Answer 7

the question is about the graph on the interval [-2,2] and you have to break it into 2 parts [-2,1] and [1,2],
to the first part you use the first equation f(x) = x+1 . you just need 2 points (-2, -1) and (1,2) and link them to get the first part (the red part from my cheating graph)
to the second part, you use the second equation f(x) = -x +3, again, just 2 points (1,2) and (2, 1) and link them together.
You whole graph should be

Answer 8

ohhh, thank you for all your help!

Answer 9

:)