C1 graph question

coconut64

Hi just wondering why does a ii) only have one solution? I thought both lines meet at 2,0 and -2,0...

thanks.

(edited 8 years ago)

Reply 1

__Adam__

Draw the line y = -x on the graph, you should see that they only cross once.

Reply 2

JLegion

Original post by coconut64

Hi just wondering why does a ii) only have one solution? I thought both lines meet at 2,0 and -2,0...

thanks.

It will only meet the graph once.

Draw the line y=-x and you'll see it'll intersect only once; if you're struggling to visualise it, a table of values is always the best thing to do :smile:

Reply 3

TeeEm

This is a question from Solomon Papers, I think.

Reply 4

Zacken

Original post by coconut64

Hi just wondering why does a ii) only have one solution? I thought both lines meet at 2,0 and -2,0...

thanks.

Look at the gradient of the f(x) graph in the negative region, it goes from -2 to 0 in the x-direction and that makes it go from 0 to 2 in the y-direction, so it has a gradient of

-1

.

That's the same gradient as the line y=-x. so if you plot y=-x, it will always be two units behind (x-direction) wise f(x) in the negative region since they have the same gradient and hence increase/decrease at the same rate, but their 'starting point' is different.

Edit to add: in a more formal manner, you can define

f

piecewise such that it is

y=-x+2

for

x \leq 0

, then in the left-half plane, it will intersect

y=-x

if and only if

-x = -x + 2 \iff 2=0

. Which is false. So it won't intersect

y=-x

in the left-half plane.

(edited 8 years ago)

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C1 graph question

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