What is the tangent of f(x)=7-(16)/(sqrt(x)),\at x=4 ?

The tangent of f(x)=7-(16)/(sqrt(x)),\at x=4 is y=x-5

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tangent of f(x)=7-(16)/(sqrt(x)),\at x=4

tangent of

f (x) = 7 - \frac{16}{x}, at x = 4

y = x - 5

Solution steps

Find the tangent point:

(4, - 1)

Find the slope of

f (x) = 7 - \frac{16}{x} : \frac{df ( x )}{d x} = \frac{8}{x ^{\frac{3}{2}}}

Compute the slope of

f (x) = 7 - \frac{16}{x}

x = 4 : m = 1

Find the line with slope m=

1

and passing through

(4, - 1) : y = x - 5

y = x - 5

y = x^{2^{x}}

(x^{5} + 3 y) d x - x d y = 0

\int \frac{x ^{3}}{49 + x ^{8}} d x

\frac{d ^{2}}{d x ^{2}} (15 e^{x^{2}})

9 - x^{2}

What is the tangent of f(x)=7-(16)/(sqrt(x)),\at x=4 ?
The tangent of f(x)=7-(16)/(sqrt(x)),\at x=4 is y=x-5