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

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

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

tangent of

2 x + x^{2} - 5, at x = 1

y = 3 x - 5

Solution steps

Find the tangent point:

(1, - 2)

Find the slope of

y = 2 x + x^{2} - 5 : \frac{d y}{d x} = 2 x + \frac{1}{x}

Compute the slope of

y = 2 x + x^{2} - 5

x = 1 : m = 3

Find the line with slope m=

3

and passing through

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

y = 3 x - 5

y = \frac{x}{4 + x ^{2}}, (0, 0)

\int \frac{x ^{2} - 8}{x ^{4}} d x

\int_{0}^{1} (x^{3} - 3 x^{2}) d x

\frac{d}{d t} (4 t)

\int_{0}^{\infty} e^{- k x} d x

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