y = 9ln(x)
<span>y' = 9x^-1 =9/x</span>
y'' = -9x^-2 =-9/x^2
curvature k = |y''| / (1 + (y')^2)^(3/2)
<span>= |-9/x^2| / (1 + (9/x)^2)^(3/2)
= (9/x^2) / (1 + 81/x^2)^(3/2)
= (9/x^2) / [(1/x^3) (x^2 + 81)^(3/2)]
= 9x(x^2 + 81)^(-3/2).
To maximize the curvature, </span>
we find where k' = 0. <span>
k' = 9 * (x^2 + 81)^(-3/2) + 9x * -3x(x^2 + 81)^(-5/2)
...= 9(x^2 + 81)^(-5/2) [(x^2 + 81) - 3x^2]
...= 9(81 - 2x^2)/(x^2 + 81)^(5/2)
Setting k' = 0 yields x = ±9/√2.
Since k' < 0 for x < -9/√2 and k' > 0 for x >
-9/√2 (and less than 9/√2),
we have a minimum at x = -9/√2.
Since k' > 0 for x < 9/√2 (and greater than 9/√2) and
k' < 0 for x > 9/√2,
we have a maximum at x = 9/√2. </span>
x=9/√2=6.36
<span>y=9 ln(x)=9ln(6.36)=16.66</span>
the
answer is
(x,y)=(6.36,16.66)
I believe the answer is A
Percent (%) = per 100
7 1/5 = 36/5 = 720/100 = 720%
The answer to the question is a
Answers: b, d and e
b.The graph has a relative minimum
d. The graph has an x intercept at 3,0
e. the graph has an y intercept at 0,-15
f(x)=(x+5)(x-3)
The given equation is in the form of f(x) = a(x-b)(x-c)
If 'a' is positive then graph has a relative minimum
If 'a' is negative then graph has a relative maximum
Here a=1 that is positive so graph has a relative minimum .
To find x intercept we set f(x) =0 and solve for x
0=(x+5)(x-3)
x+5 =0 -> x = -5 so x intercept is (-5,0)
x - 3=0 -> x= 3 so x intercept is (3,0)
To find y intercept we plug in 0 for x
y=(x+5)(x-3)
y=(0+5)(0-3) = -15
so y intercept is (0,-15)
