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

The tangent of f(x)=sqrt(x^2+3x),\at x=1 is y= 5/4 x+3/4

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

tangent of

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

y = \frac{5}{4} x + \frac{3}{4}

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

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

Compute the slope of

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

x = 1 : m = \frac{5}{4}

Find the line with slope m=

\frac{5}{4}

and passing through

(1, 2) : y = \frac{5}{4} x + \frac{3}{4}

y = \frac{5}{4} x + \frac{3}{4}

\frac{d}{d x} (4 - \frac{15}{4 x ^{\frac{1}{4}}})

\frac{d}{d x} (x^{3} + 3 x^{2} - 8)

\int 3 cos^{3} (2 x) d x

e^{x} cos (x)

\frac{d}{d t} (2)

What is the tangent of f(x)=sqrt(x^2+3x),\at x=1 ?
The tangent of f(x)=sqrt(x^2+3x),\at x=1 is y= 5/4 x+3/4