Solving for h’(-1)

[Alan Najat]

if h(x)=integral of (f’(x)x-f(x))/x^2 dx and 2h(2)=f(2)+4 and f(-1)=5 , what is the value of h’(-1) ? I have tried so hard to solve this but it always comes up dead end, some of the steps are that h(x)=f(x)/x+C and C=2 so h(x)=f(x)/x+2 and we find that h(-1)=-3 but i have no idea how to find h’(-1) since we need f’(-1) to find it. Try to hepl me solving it please, appr it

 

[Dr.Peterson]

if \(\displaystyle h(x)=\int\frac{(f’(x)x-f(x))}{x^2}dx\) and \(2h(2)=f(2)+4\) and \(f(-1)=5\), what is the value of \(h’(-1)\) ? I have tried so hard to solve this but it always comes up dead end, some of the steps are that \(h(x)=\frac{f(x)}{x}+C\) and \(C=2\) so \(h(x)=\frac{f(x)}{x}+2\) and we find that \(h(-1)=-3\) but i have no idea how to find \(h’(-1)\) since we need \(f’(-1)\) to find it. Try to help me solving it please, appr it

I added formatting to make your question easier to read above. I would have approached the work a little differently, but that doesn't help.

Since it does seem to be lacking something, could you provide an image of the problem as given to you, so we can see if there is something hidden (or misquoted)?

 

[Alan Najat]

I added formatting to make your question easier to read above. I would have approached the work a little differently, but that doesn't help.

Since it does seem to be lacking something, could you provide an image of the problem as given to you, so we can see if there is something hidden (or misquoted)?

Appreciate the formating, thanks alot, unfortunately I can’t provide the original image since it was from a telegram group and it was stricted, screenshots and saving images was disabled ,however the question wasn’t in english since i’m from kurdistan/iraq , it was in kurdish, but the question I provided is exactly the same, word to word without adding or removing a word, the teacher that had posted this problem had written “very Hard”,so i tried solving and my teacher did too, i also tried chat GPT but couldn’t reach a convincing answer, 4 options were given in the question , A)-2 ,B)2 ,C)3 ,B)-3 , and the correct answer was (-3) , h’(-1)=-3 but I don’t know how, i found out by randomly choosing one of the options,,,,,, i talked with my teacher and said a mistake might have happened during the typing in the question, i said it asks for h’(-1) and the answer is (-3) ,,, but on the other hand h(-1) is also (-3) ,,,, it might ask for h(-1) instead of h’(-1) ,,, but my teacher said we can’t decide if it is incorrect or misquoted unless the teacher that have posted the image says so,,,,, I couldn’t reach out to himself to ask him,,,, but yeah it might be (misquoted) since it clearly appears to be lacking something,,,,, but what makes me doubt that is the the teacher is very popular and the question is in a group so he also have admins so if it was a mistake there is a high chance they would notice it or get notified that there is a mistake,,,,,, i appreciate your help, thanks

 

[blamocur]

Popular teachers can make mistakes or typos too
There is no way of finding [imath]h^\prime (-1)[/imath] without any additional info. For example, consider a family of functions [imath]f=5x^{2n}[/imath], where [imath]n[/imath] is a natural number. We do get [imath]f(-1) = 5[/imath] and [imath]h(-1) = 5x^{2n-1}+2 = -3[/imath] for [imath]x=-1[/imath], but [imath]h^\prime(-1) = 10n-5[/imath], i.e. depends on [imath]n[/imath].

 

[Alan Najat]

Popular teachers can make mistakes or typos too
There is no way of finding [imath]h^\prime (-1)[/imath] without any additional info. For example, consider a family of functions [imath]f=5x^{2n}[/imath], where [imath]n[/imath] is a natural number. We do get [imath]f(-1) = 5[/imath] and [imath]h(-1) = 5x^{2n-1}+2 = -3[/imath] for [imath]x=-1[/imath], but [imath]h^\prime(-1) = 10n-5[/imath], i.e. depends on [imath]n[/imath].

yes indeed it is possible for such things to happen, and you’re absolutely right about not being able to find h’(-1) given current information, thanks a lot

 

[mario99]

Popular teachers can make mistakes or typos too
There is no way of finding [imath]h^\prime (-1)[/imath] without any additional info. For example, consider a family of functions [imath]f=5x^{2n}[/imath], where [imath]n[/imath] is a natural number. We do get [imath]f(-1) = 5[/imath] and [imath]h(-1) = 5x^{2n-1}+2 = -3[/imath] for [imath]x=-1[/imath], but [imath]h^\prime(-1) = 10n-5[/imath], i.e. depends on [imath]n[/imath].

I have a similar idea of professor blamocur, but in a different approach

We know that:

[imath]\displaystyle h'(x) = \frac{xf'(x) - f(x)}{x^2}[/imath]


[imath]\displaystyle f'(x) = \lim_{h \rightarrow 0}\frac{f(x + h) - f(x)}{h}[/imath]


[imath]\displaystyle f'(-1) = \lim_{h \rightarrow 0}\frac{f(-1 + h) - f(-1)}{h}[/imath]

We don't know the value of [imath]\displaystyle f(-1 + h)[/imath], but it is guaranteed to equal to [imath]\displaystyle 5 + v(h) [/imath], where [imath]v(h)[/imath] is also a function that we don't know. Then, we have:

[imath]\displaystyle f'(-1) = \lim_{h \rightarrow 0}\frac{5 + v(h) - 5}{h} = \lim_{h \rightarrow 0}\frac{v(h)}{h} = \lim_{h \rightarrow 0} g(h) = g(0)[/imath]


[imath]\displaystyle h'(-1) = \frac{(-1)f'(-1) - f(-1)}{(-1)^2} = -g(0) - 5[/imath]

And again the solution depends on [imath]g(0)[/imath].

 

[Alan Najat]

I have a similar idea of professor blamocur, but in a different approach

We know that:

[imath]\displaystyle h'(x) = \frac{xf'(x) - f(x)}{x^2}[/imath]


[imath]\displaystyle f'(x) = \lim_{h \rightarrow 0}\frac{f(x + h) - f(x)}{h}[/imath]


[imath]\displaystyle f'(-1) = \lim_{h \rightarrow 0}\frac{f(-1 + h) - f(-1)}{h}[/imath]

We don't know the value of [imath]\displaystyle f(-1 + h)[/imath], but it is guaranteed to equal to [imath]\displaystyle 5 + v(h) [/imath], where [imath]v(h)[/imath] is also a function that we don't know. Then, we have:

[imath]\displaystyle f'(-1) = \lim_{h \rightarrow 0}\frac{5 + v(h) - 5}{h} = \lim_{h \rightarrow 0}\frac{v(h)}{h} = \lim_{h \rightarrow 0} g(h) = g(0)[/imath]


[imath]\displaystyle h'(-1) = \frac{(-1)f'(-1) - f(-1)}{(-1)^2} = -g(0) - 5[/imath]

And again the solution depends on [imath]g(0)[/imath]great

Great approach