9514 1404 393
Answer:
maximum difference is 38 at x = -3
Step-by-step explanation:
This is nicely solved by a graphing calculator, which can plot the difference between the functions. The attached shows the maximum difference on the given interval is 38 at x = -3.
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Ordinarily, the distance between curves is measured vertically. Here that means you're interested in finding the stationary points of the difference between the functions, along with that difference at the ends of the interval. The maximum difference magnitude is what you're interested in.
h(x) = g(x) -f(x) = (2x³ +5x² -15x) -(x³ +3x² -2) = x³ +2x² -15x +2
Then the derivative is ...
h'(x) = 3x² +4x -15 = (x +3)(3x -5)
This has zeros (stationary points) at x = -3 and x = 5/3. The values of h(x) of concern are those at x=-5, -3, 5/3, 3. These are shown in the attached table.
The maximum difference between f(x) and g(x) is 38 at x = -3.
To factor quadratic equations of the form ax^2+bx+c=y, you must find two values, j and k, which satisfy two conditions.
jk=ac and j+k=b
The you replace the single linear term bx with jx and kx. Finally then you factor the first pair of terms and the second pair of terms. In this problem...
2k^2-5k-18=0
2k^2+4k-9k-18=0
2k(k+2)-9(k+2)=0
(2k-9)(k+2)=0
so k=-2 and 9/2
k=(-2, 4.5)
is this the full question?
Answer:
b is the correct anwser
Step-by-step explanation: