What is the tangent of f(x)=x^4-4x^2+7,\at x=1 ?

The tangent of f(x)=x^4-4x^2+7,\at x=1 is y=-4x+8

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tangent of f(x)=x^4-4x^2+7,\at x=1

tangent of

f (x) = x^{4} - 4 x^{2} + 7, at x = 1

y = - 4 x + 8

Solution steps

Find the tangent point:

(1, 4)

Find the slope of

f (x) = x^{4} - 4 x^{2} + 7 : \frac{df ( x )}{d x} = 4 x^{3} - 8 x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 4

Find the line with slope m=

- 4

and passing through

(1, 4) : y = - 4 x + 8

y = - 4 x + 8

\frac{d}{d x} (\frac{e ^{8 x}}{3 x ^{4}})

\int (3 x - 6 x) d x

x \to 0 lim (\frac{2 x ^{2} - x}{x})

y = (x)^{x}

f (x) = 7 sin (x)

What is the tangent of f(x)=x^4-4x^2+7,\at x=1 ?
The tangent of f(x)=x^4-4x^2+7,\at x=1 is y=-4x+8