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∵ ƒ'(x) ≥ 1 for 4 ≤ x ≤ 8

∴ ∫ ƒ'(x) dx ≥ ∫ 1 dx for 4 ≤ x ≤ 8

∴ ƒ(x) ≥ x + C for 4 ≤ x ≤ 8 ......... (1)

∴ ƒ(4) = 13 ⇒ 4 + C = 13 ⇒ C = 9

∴ from (1), ... ƒ(x) ≥ x + 9 for 4 ≤ x ≤ 8

∴ ƒ(8) ≥ 8 + 9

∴ ƒ(8) ≥ 17

∴ the smallest possible value of ƒ(8) is = 17 ... Ans.

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If the slope (f'(x)) is at least 1, that means at the very least for each time x increases by 1, f(x) is going to increase by at least one as well.

Therefore:

f(5) is at least 14

f(6) is at least 15

f(7) is at least 16

f(8) is at least 17

To do this a bit more neater mathematically, you notice that the minimum value is when is when the slope is at it's least (i.e. 1), and you know one point on the line: (4,13). So you can use this information to find the equation for the line in the slope-intercept format.

y=mx+b (m is the slope, b is the intercept with the y axis)

13=1(4) + b

13=4+b

9=b

Therefore the equation of the smallest function that suits the restrictions given is:

y=1x+9 (which can be written of course as f(x)=x+9)

Substitute 8 in:

f(8)=8+9 = 17

The smallest value of f(8) occurs when f '(x) = 1 (as the curve rises least). This means that the gradient of the curve between the points x = 4 and x=8 is one. Let the coordinates of the point at x = 8 be (8,y)

So the gradient of the interval between (4,13) and (8,y) should equal one. Using the gradient formula,

y - 13 / 8 - 4 = 1

y - 13 = 4

y = 17

So the smallest value of f(8) is 17.

f(8)=2x-1/4+3=x=1.34456