Weird result to a simple derivative question

  • Thread starter Thread starter Femme_physics
  • Start date Start date
  • Tags Tags
    Derivative Weird
Femme_physics
Gold Member
Messages
2,548
Reaction score
1
I am told to write the value of a the derivative at a certain point next to each function. And to find the point/s

f(x) = x2

f ' (x) = -6

so I figured I just do

f ' (x) = 2x

f ' (-6) = 2 x (-6) = -12

But they tell me the answer is

(-3, 9)
 
Physics news on Phys.org
The problem is that f'(x)=-6, and what you plugged in is equivalent to x=-6.

You got the derivative right, but you need to set f'(x) to -6, not x.
 
Oh... I see! thanks :)
 
Another way to look at it...
You were given value of the derivative
f'(x) = -6 (eq 1)
and you found the derivative of equation f(x)=x^2 to be
f'(x) = 2x (eq 2)

The left side of equations 1 & 2 are equal,
therefore the right sides are also,
so -6 = 2x and you can solve for x.

To find y, substitute this x into the original equation of f(x)
and you will obtain the same point you noted from the solution guide.
 
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