What is the tangent of f(x)=-x^4+4x^3+8x^2+5,\at x=2 ?

The tangent of f(x)=-x^4+4x^3+8x^2+5,\at x=2 is y=48x-43

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tangent of f(x)=-x^4+4x^3+8x^2+5,\at x=2

tangent of

f (x) = - x^{4} + 4 x^{3} + 8 x^{2} + 5, at x = 2

y = 48 x - 43

Solution steps

Find the tangent point:

(2, 53)

Find the slope of

f (x) = - x^{4} + 4 x^{3} + 8 x^{2} + 5 : \frac{df ( x )}{d x} = - 4 x^{3} + 12 x^{2} + 16 x

Compute the slope of

f (x) = - x^{4} + 4 x^{3} + 8 x^{2} + 5

x = 2 : m = 48

Find the line with slope m=

48

and passing through

(2, 53) : y = 48 x - 43

y = 48 x - 43

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\frac{d}{d x} (e^{3 x + 1})

\int \frac{1}{1 + x} d x

What is the tangent of f(x)=-x^4+4x^3+8x^2+5,\at x=2 ?
The tangent of f(x)=-x^4+4x^3+8x^2+5,\at x=2 is y=48x-43