What is the tangent of f(x)=\sqrt[4]{x}-x,\at x=1 ?

The tangent of f(x)=\sqrt[4]{x}-x,\at x=1 is y=-3/4 x+3/4

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tangent of f(x)=\sqrt[4]{x}-x,\at x=1

tangent of

y = - \frac{3}{4} x + \frac{3}{4}

Solution steps

Find the tangent point:

(1, 0)

Find the slope of

Compute the slope of

x = 1 : m = - \frac{3}{4}

Find the line with slope m=

- \frac{3}{4}

and passing through

(1, 0) : y = - \frac{3}{4} x + \frac{3}{4}

y = - \frac{3}{4} x + \frac{3}{4}

\int 5 sin (ln (x)) d x

f (x) = \frac{x}{x ^{2} + 36}

\int x^{2} (5 - x)^{4} d x

2 ln (x + 1)

n = 1 \sum \infty \frac{n !}{10 7 ^{n}}

What is the tangent of f(x)=\sqrt[4]{x}-x,\at x=1 ?
The tangent of f(x)=\sqrt[4]{x}-x,\at x=1 is y=-3/4 x+3/4