Solving for g|(-1): Finding the Answer

  • Thread starter Thread starter Dell
  • Start date Start date
Dell
Messages
555
Reaction score
0
any ideas??

given:
y=f(x)
f|(1)=4
g(x)=f(x2)

find g|(-1)
------------------------------------

i know that g(-1)=g(1) because g is a squared function of x, but that's about it,
can i say-

g|(x)=[f(x2)]|=f|(x2)*(2x)

and because x=-1, and -12=1, and i know f|(1)=4
so
g|(-1)=-8
 
Physics news on Phys.org


Dell said:
given:
y=f(x)
f|(1)=4
g(x)=f(x2)

find g|(-1)
------------------------------------

i know that g(-1)=g(1) because g is a squared function of x, but that's about it,
can i say-

g|(x)=[f(x2)]|=f|(x2)*(2x)

and because x=-1, and -12=1, and i know f|(1)=4
so
g|(-1)=-8


Yes, this is exactly right. Good job.
 


Yes, that's exactly right. Although you would be better advised to use ', a single apostrophe to indicate the derivative rather than |.
 


thanks, is there any way i would be able to solve it if i weren't given 1 as my x?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top