Given:
The function is
![f(x+2)=2x+3](https://tex.z-dn.net/?f=f%28x%2B2%29%3D2x%2B3)
To find:
The value f'(4).
Solution:
We have,
![f(x+2)=2x+3](https://tex.z-dn.net/?f=f%28x%2B2%29%3D2x%2B3)
Putting
and
we get
![f(t)=2(t-2)+3](https://tex.z-dn.net/?f=f%28t%29%3D2%28t-2%29%2B3)
![f(t)=2t-4+3](https://tex.z-dn.net/?f=f%28t%29%3D2t-4%2B3)
![f(t)=2t-1](https://tex.z-dn.net/?f=f%28t%29%3D2t-1)
Differentiate with respect to t.
![f'(t)=2(1)-0](https://tex.z-dn.net/?f=f%27%28t%29%3D2%281%29-0)
![f'(t)=2](https://tex.z-dn.net/?f=f%27%28t%29%3D2)
At
, we get
![f'(4)=2](https://tex.z-dn.net/?f=f%27%284%29%3D2)
Therefore, the value of f'(4) is 2.
---------------------------
Answer:
is -4=Y
Step-by-step explanation:
i had that problem
C= pi times d
15.3x 3.14= 48.042
X= 65
I don't really know how to solve it but that's the answer, sorry