What is the tangent of f(x)=sqrt(x^2+15),\at x=7 ?

The tangent of f(x)=sqrt(x^2+15),\at x=7 is y= 7/8 x+15/8

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tangent of f(x)=sqrt(x^2+15),\at x=7

tangent of

f (x) = x^{2} + 15, at x = 7

y = \frac{7}{8} x + \frac{15}{8}

Solution steps

Find the tangent point:

(7, 8)

Find the slope of

f (x) = x^{2} + 15 : \frac{df ( x )}{d x} = \frac{x}{x ^{2} + 15}

Compute the slope of

f (x) = x^{2} + 15

x = 7 : m = \frac{7}{8}

Find the line with slope m=

\frac{7}{8}

and passing through

(7, 8) : y = \frac{7}{8} x + \frac{15}{8}

y = \frac{7}{8} x + \frac{15}{8}

y = x

x = 1

f (x) = \frac{1}{16} x^{4} + \frac{1}{2 x ^{2}}

f (x) = tan (x)

(3, 4)

What is the tangent of f(x)=sqrt(x^2+15),\at x=7 ?
The tangent of f(x)=sqrt(x^2+15),\at x=7 is y= 7/8 x+15/8