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

The tangent of f(x)=x^4-5x^2+4,\at x=2 is y=12x-24

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

tangent of

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

y = 12 x - 24

Solution steps

Find the tangent point:

(2, 0)

Find the slope of

f (x) = x^{4} - 5 x^{2} + 4 : \frac{df ( x )}{d x} = 4 x^{3} - 10 x

Compute the slope of

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

x = 2 : m = 12

Find the line with slope m=

12

and passing through

(2, 0) : y = 12 x - 24

y = 12 x - 24

\frac{d y}{d x} - \frac{y}{x} = x e^{x}

\int \frac{x ^{2}}{x ^{2} - x + 1} d x

\frac{d}{d x} (A \cdot e^{x})

f (x) = \frac{x ^{5}}{5} - \frac{4 x ^{3}}{3} + 4 x

f (x) = 7 e^{- 0.03 x}, at x = 9

What is the tangent of f(x)=x^4-5x^2+4,\at x=2 ?
The tangent of f(x)=x^4-5x^2+4,\at x=2 is y=12x-24