What is the tangent of f(x)=x^3+6x^2+19x+15,\at x=7 ?

The tangent of f(x)=x^3+6x^2+19x+15,\at x=7 is y=250x-965

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tangent of f(x)=x^3+6x^2+19x+15,\at x=7

tangent of

f (x) = x^{3} + 6 x^{2} + 19 x + 15, at x = 7

y = 250 x - 965

Solution steps

Find the tangent point:

(7, 785)

Find the slope of

f (x) = x^{3} + 6 x^{2} + 19 x + 15 : \frac{df ( x )}{d x} = 3 x^{2} + 12 x + 19

Compute the slope of

f (x) = x^{3} + 6 x^{2} + 19 x + 15

x = 7 : m = 250

Find the line with slope m=

250

and passing through

(7, 785) : y = 250 x - 965

y = 250 x - 965

y = 10 x

(3 t^{2})^{^{'}}

\frac{d y}{d x}, y = 4^{x}

\frac{d}{d x} (ln (x^{2} + 14 x + 55))

What is the tangent of f(x)=x^3+6x^2+19x+15,\at x=7 ?
The tangent of f(x)=x^3+6x^2+19x+15,\at x=7 is y=250x-965