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$int(x+1)cdot f'(x)= x^3+x^2-x+c$ and $f(0)=frac{1}{2}$. What is $f(-1)$?

Evaluate the integral as a power series $int x^{11}cdottan^{-1}(x^2),mathrm dx$How to solve this integral $int frac{1+2x^2}{x^2(1+x^2)}dx$Question regarding $int frac{e^x}{e^x-2} ,dx$Prove $int^infty_0 bsin(frac{1}{bx})-asin(frac{1}{ax}) = -ln(frac{b}{a})$ using Frullani integralsSolving integral without simplifying equationJustification of substitution in finding indefinite integralsIndefinite integral $int frac{2^x}{2^{2x}-4}dx$Steps in Resolving the Integral: $intfrac{1}{x^3}e^frac{1}{x}dx$Solve $T'(x)=45-0.75cdot T(x)$, where $T(0)=75$.$int^infty_0 frac{dx}{x^n + 1}$, n odd

4

1

$begingroup$

$int(x+1)cdot f'(x)= x^3+x^2-x+c$ and $f(0)=frac{1}{2}$. What is $f(-1)$?

I took the derivative of both sides and then factored the quadratic equation on the right:

$$(x+1)cdot f'(x)=3x^2+2x-1$$

$$(x+1)cdot f'(x)=(3x-1)(x+1)$$

At this point, if I divide both sides of the equation by $x+1$, taking the integral becomes easy and I can figure out the answer as $3$.

Though, because we divided by $x+1$ at the beginning, we should note that $xnot = -1$. Therefore, I actually can't compute $f(-1)$.

Is there a different simple way of solving this problem?

calculus integration derivatives

share|cite|improve this question

edited Mar 10 at 4:59

rash

41412

asked Mar 10 at 4:15

Eldar RahimliEldar Rahimli

35110

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  $f(-1)$ still exists even if $f'(-1)$ does not exist.
  $endgroup$
  – Dbchatto67
  Mar 10 at 4:24
  
  
  

  
  
  
  

add a comment | 

4

1

$begingroup$

$int(x+1)cdot f'(x)= x^3+x^2-x+c$ and $f(0)=frac{1}{2}$. What is $f(-1)$?

I took the derivative of both sides and then factored the quadratic equation on the right:

$$(x+1)cdot f'(x)=3x^2+2x-1$$

$$(x+1)cdot f'(x)=(3x-1)(x+1)$$

At this point, if I divide both sides of the equation by $x+1$, taking the integral becomes easy and I can figure out the answer as $3$.

Though, because we divided by $x+1$ at the beginning, we should note that $xnot = -1$. Therefore, I actually can't compute $f(-1)$.

Is there a different simple way of solving this problem?

calculus integration derivatives

share|cite|improve this question

edited Mar 10 at 4:59

rash

41412

asked Mar 10 at 4:15

Eldar RahimliEldar Rahimli

35110

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  $f(-1)$ still exists even if $f'(-1)$ does not exist.
  $endgroup$
  – Dbchatto67
  Mar 10 at 4:24
  
  
  

  
  
  
  

add a comment | 

$begingroup$

$int(x+1)cdot f'(x)= x^3+x^2-x+c$ and $f(0)=frac{1}{2}$. What is $f(-1)$?

I took the derivative of both sides and then factored the quadratic equation on the right:

$$(x+1)cdot f'(x)=3x^2+2x-1$$

$$(x+1)cdot f'(x)=(3x-1)(x+1)$$

At this point, if I divide both sides of the equation by $x+1$, taking the integral becomes easy and I can figure out the answer as $3$.

Though, because we divided by $x+1$ at the beginning, we should note that $xnot = -1$. Therefore, I actually can't compute $f(-1)$.

Is there a different simple way of solving this problem?

calculus integration derivatives

share|cite|improve this question

edited Mar 10 at 4:59

rash

41412

asked Mar 10 at 4:15

Eldar RahimliEldar Rahimli

35110

$endgroup$

share|cite|improve this question

edited Mar 10 at 4:59

rash

41412

asked Mar 10 at 4:15

Eldar RahimliEldar Rahimli

35110

share|cite|improve this question

edited Mar 10 at 4:59

rash

41412

asked Mar 10 at 4:15

Eldar RahimliEldar Rahimli

35110

edited Mar 10 at 4:59

rash

41412

edited Mar 10 at 4:59

rash

41412

asked Mar 10 at 4:15

Eldar RahimliEldar Rahimli

35110

asked Mar 10 at 4:15

Eldar RahimliEldar Rahimli

35110

1 Answer
1

active

oldest

votes

6

$begingroup$

Since $f'(x)=3x-1$ on $(-1,0]$, by the fundamental theorem of calculus, we have
$$begin{align*}
f(0)-f(-1)&=lim_{tto -1, t>-1} f(0)-f(t)\&=lim_{tto -1, t>-1} int_t^0 (3x-1) dx\&=-2.5
end{align*}$$ hence we get $f(-1)=3$. In fact, if you have $f'(x)=3x-1$ for all $xne 1$, then mean value theorem implies the existence of $c_xin (-1,x)cup(x,-1)$ such that $frac{f(x)-f(-1)}{x-(-1)}=f'(c_x)$, hence
$$
lim_{xto -1} frac{f(x)-f(-1)}{x-(-1)}=lim_{cto -1}f'(c)=-4.
$$ So $f'(-1)=-4$ and there's no problem extending $f'(x)=3x-1$ to $x=-1$.

share|cite|improve this answer

edited Mar 10 at 4:42

answered Mar 10 at 4:25

SongSong

17.3k21246

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you please elaborate on why did you compute $f(0)-f(-1)$?
  $endgroup$
  – Eldar Rahimli
  Mar 10 at 4:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We have value of $f(0)$ and the target value if $f(-1)$, so I applied $$f(a)-f(b)=int_b^a f'(x) dx$$ to $a=0$, $bto -1$.
  $endgroup$
  – Song
  Mar 10 at 4:40
  
  
  

  
  
  
  

add a comment | 

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6

$begingroup$

Since $f'(x)=3x-1$ on $(-1,0]$, by the fundamental theorem of calculus, we have
$$begin{align*}
f(0)-f(-1)&=lim_{tto -1, t>-1} f(0)-f(t)\&=lim_{tto -1, t>-1} int_t^0 (3x-1) dx\&=-2.5
end{align*}$$ hence we get $f(-1)=3$. In fact, if you have $f'(x)=3x-1$ for all $xne 1$, then mean value theorem implies the existence of $c_xin (-1,x)cup(x,-1)$ such that $frac{f(x)-f(-1)}{x-(-1)}=f'(c_x)$, hence
$$
lim_{xto -1} frac{f(x)-f(-1)}{x-(-1)}=lim_{cto -1}f'(c)=-4.
$$ So $f'(-1)=-4$ and there's no problem extending $f'(x)=3x-1$ to $x=-1$.

share|cite|improve this answer

edited Mar 10 at 4:42

answered Mar 10 at 4:25

SongSong

17.3k21246

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you please elaborate on why did you compute $f(0)-f(-1)$?
  $endgroup$
  – Eldar Rahimli
  Mar 10 at 4:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We have value of $f(0)$ and the target value if $f(-1)$, so I applied $$f(a)-f(b)=int_b^a f'(x) dx$$ to $a=0$, $bto -1$.
  $endgroup$
  – Song
  Mar 10 at 4:40
  
  
  

  
  
  
  

add a comment | 

6

$begingroup$

Since $f'(x)=3x-1$ on $(-1,0]$, by the fundamental theorem of calculus, we have
$$begin{align*}
f(0)-f(-1)&=lim_{tto -1, t>-1} f(0)-f(t)\&=lim_{tto -1, t>-1} int_t^0 (3x-1) dx\&=-2.5
end{align*}$$ hence we get $f(-1)=3$. In fact, if you have $f'(x)=3x-1$ for all $xne 1$, then mean value theorem implies the existence of $c_xin (-1,x)cup(x,-1)$ such that $frac{f(x)-f(-1)}{x-(-1)}=f'(c_x)$, hence
$$
lim_{xto -1} frac{f(x)-f(-1)}{x-(-1)}=lim_{cto -1}f'(c)=-4.
$$ So $f'(-1)=-4$ and there's no problem extending $f'(x)=3x-1$ to $x=-1$.

share|cite|improve this answer

edited Mar 10 at 4:42

answered Mar 10 at 4:25

SongSong

17.3k21246

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you please elaborate on why did you compute $f(0)-f(-1)$?
  $endgroup$
  – Eldar Rahimli
  Mar 10 at 4:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We have value of $f(0)$ and the target value if $f(-1)$, so I applied $$f(a)-f(b)=int_b^a f'(x) dx$$ to $a=0$, $bto -1$.
  $endgroup$
  – Song
  Mar 10 at 4:40
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Since $f'(x)=3x-1$ on $(-1,0]$, by the fundamental theorem of calculus, we have
$$begin{align*}
f(0)-f(-1)&=lim_{tto -1, t>-1} f(0)-f(t)\&=lim_{tto -1, t>-1} int_t^0 (3x-1) dx\&=-2.5
end{align*}$$ hence we get $f(-1)=3$. In fact, if you have $f'(x)=3x-1$ for all $xne 1$, then mean value theorem implies the existence of $c_xin (-1,x)cup(x,-1)$ such that $frac{f(x)-f(-1)}{x-(-1)}=f'(c_x)$, hence
$$
lim_{xto -1} frac{f(x)-f(-1)}{x-(-1)}=lim_{cto -1}f'(c)=-4.
$$ So $f'(-1)=-4$ and there's no problem extending $f'(x)=3x-1$ to $x=-1$.

share|cite|improve this answer

edited Mar 10 at 4:42

answered Mar 10 at 4:25

SongSong

17.3k21246

$endgroup$

share|cite|improve this answer

edited Mar 10 at 4:42

answered Mar 10 at 4:25

SongSong

17.3k21246

answered Mar 10 at 4:25

SongSong

17.3k21246

answered Mar 10 at 4:25

SongSong

17.3k21246