Which property can be justified using the ratios in triangle XYZ
2 answers:
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Applying our power rule gets us our first derivative,
simplifying a little bit,
looking for critical points,
We can apply more factoring.
I hope this next step isn't too confusing.
We want to factor out the smallest power of x from both terms,
and also the 2 from each.
When you divide x^(-2/3) out of x^(1/3),
it leaves you with x^(3/3) or simply x.
Then apply your Zero-Factor Property,
and solve for x in each case to find your critical points.
Apply your First Derivative Test to further classify these points. You should end up finding that x=-1/2 is an relative extreme value, while x=0 is not.
Let's come back to this,
and take our second derivative.
Looking for inflection points,
Again, pulling out the smaller power of x, and fractional part,
And again, apply your Zero-Factor Property, setting each factor to zero and solving for x in each case. You should find that x=0 and x=1 are possible inflection points.
Applying your Second Derivative Test should verify that both points are in fact inflection points, locations where the function changes concavity.
simplifying a little bit,
looking for critical points,
We can apply more factoring.
I hope this next step isn't too confusing.
We want to factor out the smallest power of x from both terms,
and also the 2 from each.
When you divide x^(-2/3) out of x^(1/3),
it leaves you with x^(3/3) or simply x.
Then apply your Zero-Factor Property,
and solve for x in each case to find your critical points.
Apply your First Derivative Test to further classify these points. You should end up finding that x=-1/2 is an relative extreme value, while x=0 is not.
Let's come back to this,
and take our second derivative.
Looking for inflection points,
Again, pulling out the smaller power of x, and fractional part,
And again, apply your Zero-Factor Property, setting each factor to zero and solving for x in each case. You should find that x=0 and x=1 are possible inflection points.
Applying your Second Derivative Test should verify that both points are in fact inflection points, locations where the function changes concavity.
Answer:
see explanation
Step-by-step explanation:
the equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c the y-intercept )
• If m > 0, then the line slopes upwards from left to right ( increases )
• If m < 0, then the line slopes downwards from left to right ( decreases )
(a)
y + x = 3 ( subtract x from both sides )
y = - x + 3 ← in slope-intercept form
with m = - 1 ⇒ decreasing from left to right
(c)
y = 4x + 8 ← is in slope-intercept form
with m = 4 ⇒ increasing from left to right
the equation of a line in the form x = c represents a vertical line, where c is the value of the x-coordinates the line passes through
(b)
x - 5 = 0 ( add 5 to both sides )
x = 5 ← is a vertical line
the equation of a line in the form y = c represents a horizontal line, where c is the value of the y-coordinates the line passes through
(d)
= - 6 ( multiply both sides by 4 )
y = - 24 ← is a horizontal line