What is the tangent of f(x)=x^4-13x^2+36,\at x=1 ?

The tangent of f(x)=x^4-13x^2+36,\at x=1 is y=-22x+46

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tangent of f(x)=x^4-13x^2+36,\at x=1

tangent of

f (x) = x^{4} - 13 x^{2} + 36, at x = 1

y = - 22 x + 46

Solution steps

Find the tangent point:

(1, 24)

Find the slope of

f (x) = x^{4} - 13 x^{2} + 36 : \frac{df ( x )}{d x} = 4 x^{3} - 26 x

Compute the slope of

f (x) = x^{4} - 13 x^{2} + 36

x = 1 : m = - 22

Find the line with slope m=

- 22

and passing through

(1, 24) : y = - 22 x + 46

y = - 22 x + 46

(25 t) ((\frac{t}{25})^{2})

\frac{d}{d x} (x^{2} + \frac{1}{2 x} - 3 x)

n = 0 \sum \infty \frac{1}{2 ( 2 ) ^{n}}

\int tan^{2} (6 t) d t

x^{\frac{3}{7}}

What is the tangent of f(x)=x^4-13x^2+36,\at x=1 ?
The tangent of f(x)=x^4-13x^2+36,\at x=1 is y=-22x+46