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#### Petrus

##### Well-known member

- Feb 21, 2013

- 739

Sorry for the bad title as I did not know what to name this but this is a problem from my calculus exam which I have not decide if I shall travel 2h to get my exam and see if I got some less point then I should.. (I just got the facit for the exam and I think that I am between one higher grade that I should get)

"Supposed that \(\displaystyle f(x)\) is a function so that a tangent to function graf is \(\displaystyle x=2\) goes through \(\displaystyle (0,1)\). Supposed that also \(\displaystyle f(2)=5\) and \(\displaystyle g(x)=\frac{f(x)}{x}\)

(a) show that \(\displaystyle g'(2)=-\frac{1}{4}\)

What I did:

\(\displaystyle g'(x)=\frac{f'(x)x-f(x)}{x^2}\)

We know that \(\displaystyle g'(2)=-\frac{1}{4}\) and \(\displaystyle f(2)=5\) so I did do like this

\(\displaystyle -\frac{1}{4}=\frac{2f'(x)-5}{2^2}\) that means \(\displaystyle f'(2)=2\) and that means this is true only if \(\displaystyle f'(2)=2\)

Facit say:

they first start to calculate the tangent to the graph of \(\displaystyle f\) for \(\displaystyle x=2 \) gives of \(\displaystyle f'(2)=\frac{y-f(2)}{x-2}\)

we know that \(\displaystyle f(2)=5\) also that the point \(\displaystyle (x,y)=(0,1)\) and if we put those value on the above we get that \(\displaystyle f'(2)=2\) and then they did same as me but what so you think about how I did it?

Regards,

\(\displaystyle |\pi\rangle\)