Domain of fg(x) is determined by the domain of g(x).g(x) has domain such that x is any real number.So domain fg(x) is x such that x is any real number.
x = –6
Solution:
Given expression is ![\sqrt[3]{x-2}+5=3](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx-2%7D%2B5%3D3)
.
Step 1: Isolate the radical by subtracting 5 from both sides of the equation.
![\Rightarrow\sqrt[3]{x-2}+5-5=3-5](https://tex.z-dn.net/?f=%5CRightarrow%5Csqrt%5B3%5D%7Bx-2%7D%2B5-5%3D3-5)
![\Rightarrow\sqrt[3]{x-2}=-2](https://tex.z-dn.net/?f=%5CRightarrow%5Csqrt%5B3%5D%7Bx-2%7D%3D-2)
Step 2: Cube both sides of the equation to remove the cube root.
![\Rightarrow(\sqrt[3]{x-2})^3=(-2)^3](https://tex.z-dn.net/?f=%5CRightarrow%28%5Csqrt%5B3%5D%7Bx-2%7D%29%5E3%3D%28-2%29%5E3)
Cube and cube root get canceled in left side of the equation.
Step 3: To solve for x.
Add 2 on both sides of the equation.
Hence the solution is x = –6.
Find the area of the top ( circle) then multiply by the height.
Area = pi x r^2
Area = pi x 7^2
Area = 49pi
Volume = 49pi x 15
Volume = 735pi cubic cm.
It's false. It's a product so...
Derivative of the first TIMES the second PLUS derivative of second TIMES the first.
Derivative of the first (x^3) = 3x^2
Times the second = 3x^2 * e^x
Derivative of the second = e^x (remains unchanged)
Times the first = e^x * x^3
So the answer would be (3x^2)(e^x) + (e^x)(x^3)
which can be factorised to form x^2·e^x(3 + x)
