(x) = arcsec(x) − 8x
f'(x) = d/dx( arcsec(x) −
8x )
<span> 1/xsqrt( x^2 - 1) - 8</span>
f'(x) = 0
1/xsqrt( x^2 - 1) - 8 = 0
8 x sqrt (x^2-1) = 1
<span> ( 8 x sqrt (x^2-1) )^2 = 1</span>
64 x^2 ( x^2 - 1) = 1
64 x^4 - 64 x^2 =1
64 x^4 - 64 x^2 - 1 = 0
x = 1.00766 , - 1.00766
<span> x = - 1.00766</span>
f(- 1.00766) = arcsec(-
1.00766) − 8( - 1.00766)
f( - 1.00766 ) = 11.07949
x = 1.00766
f(1.00766) =
arcsec(1.00766) − 8( 1.00766)
f(1.00766 ) = -7.93790
relative maximum (x, y) =
(- 1.00766 , 11.07949 ) relative minimum (x, y) = ( 1.00766 ,
-7.93790 )
3/4+7/9
LCM=36
27/36+28/36
55/36
1 whole number 19/36
Or 1.527777777777778
1/8 is the answer to the question
Answer:
Step-by-step explanation:
The first parabola has vertex (-1, 0) and y-intercept (0, 1).
We plug these values into the given vertex form equation of a parabola:
y - k = a(x - h)^2 becomes
y - 0 = a(x + 1)^2
Next, we subst. the coordinates of the y-intercept (0, 1) into the above, obtaining:
1 = a(0 + 1)^2, and from this we know that a = 1. Thus, the equation of the first parabola is
y = (x + 1)^2
Second parabola: We follow essentially the same approach. Identify the vertex and the two horizontal intercepts. They are:
vertex: (1, 4)
x-intercepts: (-1, 0) and (3, 0)
Subbing these values into y - k = a(x - h)^2, we obtain:
0 - 4 = a(3 - 1)^2, or
-4 = a(2)². This yields a = -1.
Then the desired equation of the parabola is
y - 4 = -(x - 1)^2
Answer:
The solution to the system is (-2,6)
Step-by-step explanation:
I said the solution to the system is (-2,6) because that is where both lines cross, meaning that's the solution.
Hope this is hepful.
