What is the tangent of f(x)=x^2-2,\at x=4 ?

The tangent of f(x)=x^2-2,\at x=4 is y=8x-18

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tangent of f(x)=x^2-2,\at x=4

tangent of

f (x) = x^{2} - 2, at x = 4

y = 8 x - 18

Solution steps

Find the tangent point:

(4, 14)

Find the slope of

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

Compute the slope of

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

x = 4 : m = 8

Find the line with slope m=

8

and passing through

(4, 14) : y = 8 x - 18

y = 8 x - 18

f (x) = ln (x + ln (x))

\frac{x}{25 + x ^{2}}

\int \frac{x ^{2}}{x ^{6} - 3} d x

\frac{d y}{d x} y = arccos (8 x^{2})

What is the tangent of f(x)=x^2-2,\at x=4 ?
The tangent of f(x)=x^2-2,\at x=4 is y=8x-18