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`f(x) = -4x^2 + 24x+ 2`
a) Critical values are the values where the derivative is zero and does not exist. The first derivative of the function is
f'(x) = -8x + 24
This exists everywhere and is 0 when -8x + 24 = 0.
-8x = -24
x = 3
The critical value is 3.
b) The function is increasing when the first derivative is positive. This happens when -8x + 24 >0
From here -8x > -24. Divide by -8, and change the sign of inequality:
x < 3.
In interval notation, it is written as `(-oo, 3)` .
The interval when function is decreasing can be found in a similar way. The function is decreasing when the fist derivative is negative:
-8x + 24 <0
-8x < -24
x > 3
The function is decreasing on the interval
`(3, oo)` .
c) According to the First Derivative Test, if the function's derivative changes sign at a critical value, there is a local (relative) extremum at this value. The given function has one critical value x = 3, where f'(x) changes sign from positive to negative. This means f(x) has a relative maximum at x = 3.
At x = 3, `y = f(3) = -4*3^2 + 24*3 + 2 = 38` .
The relative maximum is (3, 38).
The relative minimum does not exist.
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