Example 1
Write y = x2 + 4x + 1 using function notation and evaluate the function at x = 3.
Solution
Given, y = x2 + 4x + 1
By applying function notation, we get
f(x) = x2 + 4x + 1
Evaluation:
Substitute x with 3
f (3) = 32 + 4 × 3 + 1 = 9 + 12 + 1 = 22
Example 2
Evaluate the function f(x) = 3(2x+1) when x = 4.
Solution
Plug x = 4 in the function f(x).
f (4) = 3[2(4) + 1]
f (4) = 3[8 + 1]
f (4) = 3 x 9
f (4) = 27
Example 3
Write the function y = 2x2 + 4x – 3 in function notation and find f (2a + 3).
Solution
y = 2x2 + 4x – 3 ⟹ f (x) = 2x2 + 4x – 3
Substitute x with (2a + 3).
f (2a + 3) = 2(2a + 3)2 + 4(2a + 3) – 3
= 2(4a2 + 12a + 9) + 8a + 12 – 3
= 8a2 + 24a + 18 + 8a + 12 – 3
= 8a2 + 32a + 27
Answer:
#1 is 3x + 4 =-17 #2 is 3x =-21 #3 is x = -7 #4 is 2x + x +4 = -17
Step-by-step explanation:
#1 gives you the problem ready to be worked out. #2 all you need to do to finish it is divide both sides by 3. #3 the answer is already there. #4 You need to combine the like terms then solve the problem.
Answer:
tengo
Step-by-step explanation:
x = –6
Solution:
Given expression is
.
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.
13/32=0.40625 that would be the answer to the problem