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Given the function, the output value of f(0) is 5 and the value of x when f(x)=0 are 5 and 1.
Given the function in the question;
To determine the output value of f(0), replace all the occurrence of x with 0 in the function and simplify.
f(x) = x² - 6x + 5
f(0) = (0)² - (0)x + 5
f(0) = 0 - 0 + 5
f(0) = 5
To determine the value of x when f(x) = 0, replace f(x) with 0 in the function and simplify.
f(x) = x² - 6x + 5
0 = x² - 6x + 5
x² - 6x + 5 = 0
Factor x² - 6x + 5 using the AC method.
(x-5)(x-1) = 0
x = 5, x = 1
x = 5, 1
Therefore, the output value of f(0) is 5 and the value of x when f(x)=0 are 5 and 1.
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