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

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

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

tangent of

f (x) = \frac{1}{1 + x ^{2}}, (- 1, \frac{1}{2})

y = \frac{1}{2} x + 1

Solution steps

Find the slope of

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

Compute the slope of

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

(- 1, \frac{1}{2}) : m = \frac{1}{2}

Find the line with slope m=

\frac{1}{2}

and passing through

(- 1, \frac{1}{2}) : y = \frac{1}{2} x + 1

y = \frac{1}{2} x + 1

f (x) = \frac{sin ( x )}{2 x}

\int_{- π}^{π} sin (x) sin (n x) d x

\frac{1}{3 + 5 x}

y = x^{\frac{5}{2}} e^{x}

f (x) = \frac{ln ( x )}{2 x}

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