What is the tangent of f(x)=x^2+1,(0,1) ?

The tangent of f(x)=x^2+1,(0,1) is y=1

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tangent of f(x)=x^2+1,(0,1)

tangent of

f (x) = x^{2} + 1, (0, 1)

y = 1

Solution steps

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

(0, 1) : m = 0

Find the line with slope m=

0

and passing through

(0, 1) : y = 1

y = 1

n = 5 \sum \infty \frac{3 ^{n}}{1 2 ^{n}}

y = x^{2} - 5

x \to - 1 lim (4 x^{2} - 2 x - 6)

\frac{d}{d x} (\frac{1 - 4 ln ( x )}{x ^{5}})

sin (2 π t)

What is the tangent of f(x)=x^2+1,(0,1) ?
The tangent of f(x)=x^2+1,(0,1) is y=1