I haven't learned antiderivitives yet but I can try to logic it
<span>First find f′ and then find f. f′′(x)=3x^3+6x^2−x+2, f′(1)=9, f(1)=−7.
we reverse chain rule
3x^3, we know that it was a 4th degree thing, and the coefient is 3, so
4*what=3?, answer is 3/4
3/4x^4
6x^2
we know it was x^3, and the coefient is now 6 so
3*what=6? what=2
2x^3
-1x, the power was 2 and coefient is now -1, so
2 times what=-1?, -1/2
-1/2x^2
2, that is from 2x
so
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<span>3/4x^4+2x^3-1/2x^2+2x=f'(x)
test x=1
(3/4)(1)+2(1)-(1/2)(1)+2(1)=
3/4+2-2/4+2=
4 and 1/4 we need to get it to 9
4 and 1/4 +what=9
answer is 4 and 3/4
so we add that to the end since it will become 0 from derivitive
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<span>f'(x)=3/4x^4+2x^3-1/2x^2+2x+4 and 3/4
now reverse drivitive again
3/4x^4
exponent is 5 and coefient is 3/4
5 times what=3/4? answer is 3/20
3/20x^5
2x^3
exponent should be 4 and coefient is 2
4 times what=2? answer is 1/2
1/2x^4
-1/2x^2
exponent should be 3 and coefient is -1/2
3 times what=-1/2? answer is -1/6
-1/6x^3
2x
exponent should be 2 and coefient is 2
2 times what=2? answer is 1
1x^2
4 and 3/4 turns to (4 and 3/4)x
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<span>f(x)=3/20x^5+1/2x^4-1/6x^3+x^2+(4 and 3/4)x
try evaluating it for x=1
f(1)=(3/20)+(10/20)-(10/50)+1+(19/4)
f(1)=6 and 7/30
what do we add to get -7
-13 and 7/30
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<span><span>f(x)=3/20x^5+1/2x^4-1/6x^3+x^2+(4 and 3/4)x-13 and 7/30
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</span>ANSWER
<span>f'(x)=3/4x^4+2x^3-1/2x^2+2x+19/4
</span><span>f(x)=3/20x^5+1/2x^4-1/6x^3+x^2+19/4x-187/30</span>
its lonmg.answer
Step-by-step explanation:
Answer:
1 is the answer
Step-by-step explanation:
-5-(-6) = x
-5 + 6 = x
1 = x
Part A: As we have learned when given a pair of equations, we must always find a Point of Intersection....or, in other words...the solution of the lines given to us. As we know there can be one solution, in this case we view the graph; see where they intersect, and realize that the solution to these lines (Point of Intersection) is at (4,4).
Thus, your answer.
Only 2 forces are out of balance - the 22N and the 14N
So the resultant force is 22-14 = 8N.
