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

The tangent of f(x)=3x^2+x-2,\at x=1 is y=7x-5

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

tangent of

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

y = 7 x - 5

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

f (x) = 3 x^{2} + x - 2 : \frac{df ( x )}{d x} = 6 x + 1

Compute the slope of

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

x = 1 : m = 7

Find the line with slope m=

7

and passing through

(1, 2) : y = 7 x - 5

y = 7 x - 5

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

\frac{\partial}{\partial x} (7 e^{x y} ln (x^{2} + y^{2}))

\frac{d}{d x} (x^{- \frac{5}{3}})

\frac{1}{2 s + 4}

f (x) = x^{2} - 3, g (x) = 5 - x^{2}

What is the tangent of f(x)=3x^2+x-2,\at x=1 ?
The tangent of f(x)=3x^2+x-2,\at x=1 is y=7x-5