Step-by-step explanation:
by using y=uv derivative formula and stationary mean y' = 0
y' = 3(x–2)⁴ + 3(3x–1)(x–2)³ = 0
cancel 3 and factorize
(x–2+3x–1)(x–2)³ = 0
x = ¾ or x = 2
we got point
and (2,0)
Answer:
![h(x)\leq 5](https://tex.z-dn.net/?f=h%28x%29%5Cleq%205)
Step-by-step explanation:
<u>Piecewise Function</u>
The given function has a point where its behavior changes. At the left side of x=3 the function is
![h(x)=x+2](https://tex.z-dn.net/?f=h%28x%29%3Dx%2B2)
While at the right side the function is
![h(x)=-x+8](https://tex.z-dn.net/?f=h%28x%29%3D-x%2B8)
We need to find the range of h, i.e. the values h takes when x moves along its domain. Let's analyze what values takes h when x decreases without limit.
![\lim\limits_{x \rightarrow -\infty }(x+2)=-\infty](https://tex.z-dn.net/?f=%5Clim%5Climits_%7Bx%20%5Crightarrow%20-%5Cinfty%20%7D%28x%2B2%29%3D-%5Cinfty)
When x=3
![h(3)=-3+8=5](https://tex.z-dn.net/?f=h%283%29%3D-3%2B8%3D5)
When x increses without limit
![\lim\limits_{x \rightarrow \infty }(-x+8)=-\infty](https://tex.z-dn.net/?f=%5Clim%5Climits_%7Bx%20%5Crightarrow%20%5Cinfty%20%7D%28-x%2B8%29%3D-%5Cinfty)
As shown, the values of h run from
to 5, i.e.
![Range \ h=(-\infty , 5]](https://tex.z-dn.net/?f=Range%20%5C%20h%3D%28-%5Cinfty%20%2C%205%5D)
Or, equivalently
![\boxed{h(x)\leq 5}](https://tex.z-dn.net/?f=%5Cboxed%7Bh%28x%29%5Cleq%205%7D)
Ivan will need 24 cans. 56/10 = 5.6. Each can is 5.6 cm tall. 162.4-28 =134.4
134.4/5.6 = 24
Answer:
32
Step-by-step explanation:
Answer:
![x = \frac{-5+/-\sqrt{57} }{2}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B-5%2B%2F-%5Csqrt%7B57%7D%20%7D%7B2%7D)
Step-by-step explanation:
To solve the quadratic equation, rearrange it into
.
![x^2=-5x+8\\x^2 + 5x-8](https://tex.z-dn.net/?f=x%5E2%3D-5x%2B8%5C%5Cx%5E2%20%2B%205x-8)
This quadratic is not factorable and to solve needs the quadratic formula.
The quadratic formula is
. Here a=1, b=5, and c=-8.
Substitute and you'll have: