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

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

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

tangent of

f (x) = x^{2} + 3 x - 1, at x = 1

y = 5 x - 2

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

f (x) = x^{2} + 3 x - 1 : \frac{df ( x )}{d x} = 2 x + 3

Compute the slope of

f (x) = x^{2} + 3 x - 1

x = 1 : m = 5

Find the line with slope m=

5

and passing through

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

y = 5 x - 2

y^{^{''}} - 4 = 0

\frac{d}{d x} ((2 x + 3) (3 x - 2))

6 y^{^{'}} - y = 9 x^{- 3 x}, y (0) = 4

\frac{2 x + 4}{3 x + 7}, at - 2

\frac{d}{d x} (\frac{sin ( x y )}{e ^{x} - y ^{2}})

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