Question

find the tangent to curve y=3x^3-4x^2+5x+6 at (1,1)

Answer 1

Answer:

Slope of the tangent at any point to the curve:[math]y=3x^2-x+1[/math]

[math]\frac{dy}{dx}=\frac{d}{dx}(3x^2-x+1)=6x-1[/math]

hence, the slope of tangent at point [math](1, 3)[/math] is [math]\left(\frac{dy}{dx}\right)_{x=1, y=3}=6\cdot 1-1=5[/math]

hence the equation of tangent at point [math](1, 3)[/math] having slope [math]5[/math] is given by [math]y-y_1=m(x-x_1)[/math] as follows

Step-by-step explanation:

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