What is the tangent of f(x)=x^4-8x^2+5x+4,\at x=1 ?

The tangent of f(x)=x^4-8x^2+5x+4,\at x=1 is y=-7x+9

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tangent of f(x)=x^4-8x^2+5x+4,\at x=1

tangent of

f (x) = x^{4} - 8 x^{2} + 5 x + 4, at x = 1

y = - 7 x + 9

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

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

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 = - 7

Find the line with slope m=

- 7

and passing through

(1, 2) : y = - 7 x + 9

y = - 7 x + 9

\frac{3 s + 4}{s ^{2} + 10 s + 21}

f^{^{'}} (x) = 16 x

e^{(- 2 t)}

\int - \frac{3}{x ^{4}} d x

\int \frac{x ^{2} + 3 x + 5}{x ^{3} + 8} d x

What is the tangent of f(x)=x^4-8x^2+5x+4,\at x=1 ?
The tangent of f(x)=x^4-8x^2+5x+4,\at x=1 is y=-7x+9