Question

Q.12
12. If f(x) = 4x - x2, then find f (a+1)-f (a-1).
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Q.13​

Answer 1

`f(a+1) -f (a-1) = 8 -4a`

Explanation:

Given

`f(x) = 4x - x^2`

Required:

`f(a+1) -f (a-1)`

First, we solve for
`f(a+1)`

This is solved by substituting a + 1 for x in f(x)

`f(x) = 4x - x^2` becomes

`f(a + 1) = 4(a + 1) - (a+1)^2`

Open bracket

`f(a + 1) = 4a + 4 - (a+1)(a+1)`

`f(a + 1) = 4a + 4 - (a^2 + a + a+1)`

`f(a + 1) = 4a + 4 - (a^2 + 2a+1)`

Open bracket

`f(a + 1) = 4a + 4 -a^2 - 2a -1`

Collect like terms

`f(a + 1) = 4 - 1 + 4a - 2a -a^2`

`f(a + 1) = 3 + 2a -a^2`

Solving for
`f(a-1)`

This is solved by substituting a - 1 for x in f(x)

`f(x) = 4x - x^2` becomes

`f(a - 1) = 4(a - 1) - (a-1)^2`

Open bracket

`f(a - 1) = 4a - 4 - (a-1)(a-1)`

`f(a - 1) = 4a - 4 - (a^2 - a - a+1)`

`f(a - 1) = 4a - 4 - (a^2 - 2a+1)`

Open bracket

`f(a - 1) = 4a - 4 -a^2 + 2a- 1`

Collect like terms

`f(a - 1) = -4 - 1 + 4a + 2a -a^2`

`f(a - 1) = -5 + 6a -a^2`

`f(a+1) -f (a-1)` becomes

`f(a+1) -f (a-1) = 3 + 2a -a^2 - (-5 + 6a -a^2)`

Open bracket

`f(a+1) -f (a-1) = 3 + 2a -a^2 +5 - 6a + a^2`

Collect like terms

`f(a+1) -f (a-1) = 3 +5 + 2a - 6a -a^2 + a^2`

`f(a+1) -f (a-1) = 8 -4a`