What is the tangent of f(x)=sqrt(3x+4),\at x=4 ?

The tangent of f(x)=sqrt(3x+4),\at x=4 is y= 3/8 x+5/2

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tangent of f(x)=sqrt(3x+4),\at x=4

tangent of

f (x) = 3 x + 4, at x = 4

y = \frac{3}{8} x + \frac{5}{2}

Solution steps

Find the tangent point:

(4, 4)

Find the slope of

f (x) = 3 x + 4 : \frac{df ( x )}{d x} = \frac{3}{2 3 x + 4}

Compute the slope of

f (x) = 3 x + 4

x = 4 : m = \frac{3}{8}

Find the line with slope m=

\frac{3}{8}

and passing through

(4, 4) : y = \frac{3}{8} x + \frac{5}{2}

y = \frac{3}{8} x + \frac{5}{2}

f (x) = x^{4} - 32 x^{2} + 256, at (1,)

\int cos (θ) d θ

\int \frac{n}{n + 1} d n

\int (3 x + \frac{10}{x ^{6}}) d x

f (x) = 3 x^{\frac{1}{2}} + x^{\frac{3}{2}}, at x = 9

What is the tangent of f(x)=sqrt(3x+4),\at x=4 ?
The tangent of f(x)=sqrt(3x+4),\at x=4 is y= 3/8 x+5/2