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

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

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

tangent of

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

y = 4 x - 3

Solution steps

Find the tangent point:

(2, 5)

Find the slope of

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

Compute the slope of

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

x = 2 : m = 4

Find the line with slope m=

4

and passing through

(2, 5) : y = 4 x - 3

y = 4 x - 3

\int 12 - x d x

t \to - 1 lim (1 - 2 t)

\frac{6}{s ^{2} ( s ^{2} + 8 s + 1 )}

\int 3 x^{2} - 4 x + 7 d x

y = 2 x - x, y = 0

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