Please help me with geometry!
1 answer:
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Move everybody to one side
minus 5x+2 both sides
x^2-5x-2=0
use quadratic formula where
if you have
ax^2+bx+c=0
x=
1x^2-5x-2=0
a=1
b=-5
c=-2
x=
x=
x=
aprox
x=5.372281323 or -0.3722813233
minus 5x+2 both sides
x^2-5x-2=0
use quadratic formula where
if you have
ax^2+bx+c=0
x=
1x^2-5x-2=0
a=1
b=-5
c=-2
x=
x=
x=
aprox
x=5.372281323 or -0.3722813233
By definition of absolute value, you have
or more simply,
On their own, each piece is differentiable over their respective domains, except at the point where they split off.
For <em>x</em> > -1, we have
(<em>x</em> + 1)<em>'</em> = 1
while for <em>x</em> < -1,
(-<em>x</em> - 1)<em>'</em> = -1
More concisely,
Note the strict inequalities in the definition of <em>f '(x)</em>.
In order for <em>f(x)</em> to be differentiable at <em>x</em> = -1, the derivative <em>f '(x)</em> must be continuous at <em>x</em> = -1. But this is not the case, because the limits from either side of <em>x</em> = -1 for the derivative do not match:
All this to say that <em>f(x)</em> is differentiable everywhere on its domain, <em>except</em> at the point <em>x</em> = -1.