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

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

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

tangent of

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

y = \frac{7}{2} x - \frac{3}{2}

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

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

Compute the slope of

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

x = 1 : m = \frac{7}{2}

Find the line with slope m=

\frac{7}{2}

and passing through

(1, 2) : y = \frac{7}{2} x - \frac{3}{2}

y = \frac{7}{2} x - \frac{3}{2}

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

\int_{0}^{\frac{π}{2}} 2 cos^{2} (x) d x

x \to - 0 lim (\frac{6 x}{x ^{4}})

\frac{- 4 e ^{(- s)} \cdot ( e ^{3} )}{( ( s - 3 ) ( s - 4 ) ( s + 1 ) )}

y^{^{'}} + y = e^{- 2 t}

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