What's the difference between the maximum value of y=-x^4-2x^3+5 and the minimum value of y=x^4+x^2-4? (a) 1 (b) 1.5 (c) 2.7 (d) 9 (e) 10.7

You need to solve the equation `(dy)/(dx) = 0`  to find the maximum value of `y = -x^4-2x^3+5`  such that:

`-4x^3 - 6x^2 = 0`

You need to divide by -2 such that:

`2x^3 + 3x^2 = 0`

You need to factor out `x^2`  such that:

`x^2(2x + 3) =...

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You need to solve the equation `(dy)/(dx) = 0`  to find the maximum value of `y = -x^4-2x^3+5`  such that:

`-4x^3 - 6x^2 = 0`

You need to divide by -2 such that:

`2x^3 + 3x^2 = 0`

You need to factor out `x^2`  such that:

`x^2(2x + 3) = 0`

`x^2 = 0 =gt x = 0`

`2x + 3 = 0 =gt 2x = -3 =gt x = -3/2`

You need to select for x a value larger than 0, such that:

`f'(1) = -10lt0`

You need to select for x a value between `-3/2`  and 0 such that:

`f'(-1) = 1gt0`

You need to select for x a value smaller than `-3/2 ` such that:

`f'(-2) = 4(-4+3) = -4lt0`

Hence, the function `y = -x^4-2x^3+5`  reaches  the maximum value at `x = 0` .

You need to solve the equation `(dy)/(dx) = 0 ` to find the maximum value of `y=x^4+x^2-4`  such that:

`4x^3 + 2x = 0`

`2x^3 + x = `  0

`x(2x^2 + 1) = 0`

`x=0`

`x^2 = -1/2 =gt x !in R`

Hence, the function `y=x^4+x^2-4`  reaches its minimum value at `x=0` .

Hence, evaluating the difference between maximum value of`y = -x^4-2x^3+5`  and the minimum value of `y=x^4+x^2-4`  yields 0.

Notice that you may get the answer b) if you evaluate the difference between the minimum value of `y=x^4+x^2-4`  and the minimum value of `y = -x^4-2x^3+5`  such that: `0 - (-3/2) = 0+1.5 = 1.5.`

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