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

The tangent of f(x)=x^2+1 is y=2a_{0}x-a_{0}^2+1

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

tangent of

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

y = 2 a_{0} x - a_{0}^{2} + 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} + 1)

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 = a_{0} : m = 2 a_{0}

Find the line with slope m=

2 a_{0}

and passing through

(a_{0}, a_{0}^{2} + 1) : y = 2 a_{0} x - a_{0}^{2} + 1

y = 2 a_{0} x - a_{0}^{2} + 1

7 y^{^{''}} + 5 y^{^{'}} + 7 y = 0

h \to 0 lim (\frac{\frac{4}{2 + h} - 2}{h})

x = 1 - y^{2}, - x - 1

\frac{d}{d x} (2 p (x) x)

arctan (\frac{x}{3})

What is the tangent of f(x)=x^2+1 ?
The tangent of f(x)=x^2+1 is y=2a_{0}x-a_{0}^2+1