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

The tangent of f(x)=x^2+2,\at x=3 is y=6x-7

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

tangent of

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

y = 6 x - 7

Solution steps

Find the tangent point:

(3, 11)

Find the slope of

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

Compute the slope of

f (x) = x^{2} + 2

x = 3 : m = 6

Find the line with slope m=

6

and passing through

(3, 11) : y = 6 x - 7

y = 6 x - 7

\frac{d}{d x} (4 x arcsin (x))

\frac{d}{d x} (\frac{ln ( 1 )}{x})

\frac{\partial}{\partial x} (2 e^{y} sin (- 3 x) - 4 y + x y)

7 (ln (y)) y^{^{'}} - x^{3} y = 0

\frac{e ^{- 2 s}}{s ( s + 1.1 )}

What is the tangent of f(x)=x^2+2,\at x=3 ?
The tangent of f(x)=x^2+2,\at x=3 is y=6x-7