`f''(x) = 8x^3 + 5, f(1) = 0, f'(1) = 8` Find `f`.

Textbook Question

Chapter 4, 4.9 - Problem 40 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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`f''(x)=8x^3+5`

`f'(x)=int(8x^3+5)dx`

`f'(x)=8(x^4/4)+5x+c_1`

`f'(x)=2x^4+5x+c_1`

Now let's find constant c_1 , given f'(1)=8

`f'(1)=8=2(1)^4+5(1)+c_1`

`8=2+5+c_1`

`c_1=1`

`:.f'(x)=2x^4+5x+1`

`f(x)=int(2x^4+5x+1)dx`

`f(x)=2(x^5/5)+5(x^2/2)+x+c_2`

Now let's find constant c_2 , given f(1)=0

`f(1)=0=2(1^5/5)+5(1^2/2)+1+c_2`

`0=2/5+5/2+1+c_2`

`c_2=-(2/5+5/2+1)=-(2*2+5*5+10)/10=-39/10`

`:.f(x)=(2x^5)/5+(5x^2)/2+x-39/10`

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scisser | (Level 3) Honors

Posted on

Find the anti derivatives and solve for the constants.

`int(8x^3+5)=f'(x)=2x^4+5x+C`

`f'(1)=8` , therefore,

`2(1)^4+5(1)+C=8`

`C=1`

Thus, `f'(x)=2x^4+5x+1`

Do the same thing again

`int(2x^4+5x+1)=f(x)=(2/5)x^5+(5/2)x^2+x+C`

Plug in the point` f(1)=0`

`0=(2/5)(1)^5+(5/2)(1)^2+(1)+C`

`C=-39/10`

Thus,

`f(x)=2/5x^5+5/2x^2+x-39/10`

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