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I got absolute minimum wrong

9514 1404 393

Step-by-step explanation:

The absolute extremes will be at one of two locations:

Here, the absolute maximum occurs where the derivative is zero (and the second derivative is negative).

  f'(x) = 3x^2·e^-x -x^3·e^-x = x^2·e^-x·(3 -x)   ⇒   f'(3) = 0

The absolute minimum occurs at the left end of the interval, where both x^3 and e^-x have large magnitude.

The minimum is ...

  (-4)^3×e^-(-4) ≈ -3494.21 . . . . six significant figures

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A graph can show you the function tends toward -∞ as x goes to large negative values. Hence, the absolute minimum on an interval beginning in the left half-plane will be at the left end of the interval.

step-by-step explanation:

i found the koficent of the linear function and i foun c acordind to the strict formula y=kx+c and then i took 2 as an x and replacet it on the formula i found and then foun the y

1. 3. can also be expressed as (22/7)

2. -4.5

3. 3.6

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