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

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

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

tangent of

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

y = \frac{5}{2} x - \frac{1}{2}

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

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

Compute the slope of

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

x = 1 : m = \frac{5}{2}

Find the line with slope m=

\frac{5}{2}

and passing through

(1, 2) : y = \frac{5}{2} x - \frac{1}{2}

y = \frac{5}{2} x - \frac{1}{2}

\frac{1 - 6 t}{5 + t}

\int 11 tan^{4} (x se) c^{6} x d x

x, x^{2}

\frac{d}{d x} (1 - e^{3 x})

\int_{0}^{2} x^{2} (1 - 0.5 x) d x

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