Solving for k: -1/2 | Curve Slope, Value of k, and Equations Explained

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Homework Statement


If the slope of the curve y=−2kx^3+4kx at x=3 is 25, what is the value of k?

Homework Equations


f(x)=xn
f'(x)=nxn-1

The Attempt at a Solution


y=f(x)

f(x)=-2kx3+4kx
f'(x)=-6kx2+4k
x=3
f'(3)=-50k

-50k = 25
-k=1/2
k=-1/2

Did I do it right?
 
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caters said:

Homework Statement


If the slope of the curve y=−2kx^3+4kx at x=3 is 25, what is the value of k?

Homework Equations


f(x)=xn
f'(x)=nxn-1

The Attempt at a Solution


y=f(x)

f(x)=-2kx3+4kx
f'(x)=-6kx2+4k
x=3
f'(3)=-50k

-50k = 25
-k=1/2
k=-1/2

Did I do it right?

Yes.
 
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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...
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