Equation of Tangent: y = -7x + 5 @ (1,1)

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Determine the equation 1 the tangent to the given function, at the given point.

y = (3x^-2 - 2x^3) , @ (1,1)
 
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pretty sure you need to take the derivative of that. The dy/dx (derivative) you get out of that after plugging everything in is your slope of the tangent. Then just solve for your y-intercept.

*edit* The other questions you just posted are really the same idea as this.
 
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Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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