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

The tangent of f(x)=x^2+1,\at x=3 is y=6x-8

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

tangent of

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

y = 6 x - 8

Solution steps

Find the tangent point:

(3, 10)

Find the slope of

f (x) = x^{2} + 1 : \frac{df ( x )}{d x} = 2 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 = 6

Find the line with slope m=

6

and passing through

(3, 10) : y = 6 x - 8

y = 6 x - 8

x \to 3 lim (\frac{5}{( x - 3 ) ^{2}})

\frac{d y}{d x} = \frac{1}{x} (y^{2} + y), y (1) = 0.5

\int x^{3} ln (x) d x

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

(x + 1) y^{^{'}} + (x + 2) y = 2 x e^{- x}

What is the tangent of f(x)=x^2+1,\at x=3 ?
The tangent of f(x)=x^2+1,\at x=3 is y=6x-8