What is the tangent of f(x)=x^2+x,\at x=1,x=5 ?

The tangent of f(x)=x^2+x,\at x=1,x=5 is y=3x-1

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tangent of f(x)=x^2+x,\at x=1,x=5

tangent of

f (x) = x^{2} + x, at x = 1, x = 5

y = 3 x - 1

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

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

Compute the slope of

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

x = 1 : m = 3

Find the line with slope m=

3

and passing through

(1, 2) : y = 3 x - 1

y = 3 x - 1

(x^{3})^{^{'}}

\frac{d y}{d x} = \frac{( y ^{2} )}{x}

\frac{d}{d x} (\frac{1}{2})^{s i n^{2} (x)}

\int \frac{4 x - 1}{x ^{3} + 2 x ^{2} + x} d x

\frac{d}{d x} (\frac{1}{( 2 - x )})

What is the tangent of f(x)=x^2+x,\at x=1,x=5 ?
The tangent of f(x)=x^2+x,\at x=1,x=5 is y=3x-1