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 </span> <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
</span> <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
</span> <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 </span> <span><span>f(x)=3/20x^5+1/2x^4-1/6x^3+x^2+(4 and 3/4)x-13 and 7/30
