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

The tangent of f(x)=sqrt(x+3),(1,2) is y= 1/4 x+7/4

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

tangent of

f (x) = x + 3, (1, 2)

y = \frac{1}{4} x + \frac{7}{4}

Solution steps

Find the slope of

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

Compute the slope of

f (x) = x + 3

(1, 2) : m = \frac{1}{4}

Find the line with slope m=

\frac{1}{4}

and passing through

(1, 2) : y = \frac{1}{4} x + \frac{7}{4}

y = \frac{1}{4} x + \frac{7}{4}

f (x) = x^{2} - 3, y = - 2 x

\int cot (36 x) d x

3 (x^{3} - x), 0, 0, 1

\int \frac{1}{x ^{4} - 10 x ^{2} + 9} d x

x \to 0 lim ((1 - 3 x)^{\frac{4}{x}})

What is the tangent of f(x)=sqrt(x+3),(1,2) ?
The tangent of f(x)=sqrt(x+3),(1,2) is y= 1/4 x+7/4