What is the tangent of f(x)=2x^2+1,(1,3) ?

The tangent of f(x)=2x^2+1,(1,3) is y=4x-1

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tangent of f(x)=2x^2+1,(1,3)

tangent of

f (x) = 2 x^{2} + 1, (1, 3)

y = 4 x - 1

Solution steps

Find the slope of

f (x) = 2 x^{2} + 1 : \frac{df ( x )}{d x} = 4 x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 4

Find the line with slope m=

4

and passing through

(1, 3) : y = 4 x - 1

y = 4 x - 1

\frac{\partial}{\partial x} (4 cos (x y^{2}))

\int (x^{3} - \frac{1}{2 x} + 1) d x

\int - sin (- x) d x

\frac{\partial}{\partial x} (x^{3} + x)

\int_{b}^{1} 3 ln (x) d x

What is the tangent of f(x)=2x^2+1,(1,3) ?
The tangent of f(x)=2x^2+1,(1,3) is y=4x-1