Question
$y=2 x^{2}-x+1$$\frac{d y}{d x}=4 x_{1}-1$$4 x_{1}-1=3$$x_{1}=1$$y_{1}=2$
Step 1
The derivative of a function gives us the slope of the tangent line at any point. The derivative of the function is given by: \[\frac{dy}{dx}=4x-1\] Show more…
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$y=\frac{x^{2}}{4}-2$ $\frac{d y}{d x}=\frac{x}{2}$ $\left(y-\frac{x_{\perp}^{2}}{4}+2\right)=\frac{-2}{x}\left(x-x_{1}\right)$ $-1-\frac{x_{1}^{2}}{4}+2=\frac{-2}{x_{1}}\left(1-x_{1}\right)$ $\Rightarrow 2 x_{1}-\frac{x_{1} 3}{4}=-2+2 x_{1}$ $x_{1}^{3}=8$ $x_{1}=2, \quad y=-1$
$$ y=\frac{1}{2} x^{3}, \quad y=4, \quad x=0 $$
Applications of Integration
Volume: The Disk Method
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