What is the tangent of f(x)=sqrt(5x+1),\at x=7 ?

The tangent of f(x)=sqrt(5x+1),\at x=7 is y= 5/12 x+37/12

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tangent of f(x)=sqrt(5x+1),\at x=7

tangent of

f (x) = 5 x + 1, at x = 7

y = \frac{5}{12} x + \frac{37}{12}

Solution steps

Find the tangent point:

(7, 6)

Find the slope of

f (x) = 5 x + 1 : \frac{df ( x )}{d x} = \frac{5}{2 5 x + 1}

Compute the slope of

f (x) = 5 x + 1

x = 7 : m = \frac{5}{12}

Find the line with slope m=

\frac{5}{12}

and passing through

(7, 6) : y = \frac{5}{12} x + \frac{37}{12}

y = \frac{5}{12} x + \frac{37}{12}

e^{- 3 x}

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

\int cos (ln (x)) d x

\int \frac{x ^{2}}{49 + x ^{2}} d x

\int x (1 + x^{3}) d x

What is the tangent of f(x)=sqrt(5x+1),\at x=7 ?
The tangent of f(x)=sqrt(5x+1),\at x=7 is y= 5/12 x+37/12