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

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

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

tangent of

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

y = 5 x - 2

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

f (x) = 2 x^{2} + x : \frac{df ( x )}{d x} = 4 x + 1

Compute the slope of

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

x = 1 : m = 5

Find the line with slope m=

5

and passing through

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

y = 5 x - 2

\int_{0}^{1} (8 x - 1)^{2} d x

\int_{0}^{1} 7^{3 x} d x

\int \frac{( ln ( x ) - 1 ) ^{2}}{x} d x

y = \frac{x}{3} + \frac{3 x ^{2}}{2}

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