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let f:[0,10] to [1,20] be a function defined as f(x)= 60-5x/3,0≤x≤6 10, 6≤x≤7 then f is 31-3x, 7≤x≤10
Asked by msriramulu508 | 15 Mar, 2024, 08:50: AM
Expert Answer
The function is defined as:
f(x) = { {60 - 5x}/3, 0 <= x <=6 10, 6 <= x <= 7, 31 - 3x, 7 <= x <= 10
Now,
Thus, f is continuous at x = 6, 7
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