What is the tangent of f(x)=x^3-3/2 x^2+x,\at x=1 ?

The tangent of f(x)=x^3-3/2 x^2+x,\at x=1 is y=x-1/2

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tangent of f(x)=x^3-3/2 x^2+x,\at x=1

tangent of

f (x) = x^{3} - \frac{3}{2} x^{2} + x, at x = 1

y = x - \frac{1}{2}

Solution steps

Find the tangent point:

(1, \frac{1}{2})

Find the slope of

f (x) = x^{3} - \frac{3}{2} x^{2} + x : \frac{df ( x )}{d x} = 3 x^{2} - 3 x + 1

Compute the slope of

f (x) = x^{3} - \frac{3}{2} x^{2} + x

x = 1 : m = 1

Find the line with slope m=

1

and passing through

(1, \frac{1}{2}) : y = x - \frac{1}{2}

y = x - \frac{1}{2}

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

\int 4 + y^{2} d y

\int \frac{14}{( x - 1 ) ^{2} ( x - 2 ) ^{2}} d x

f (x) = arcsin (4 x)

x \to \infty lim (arctan (x^{2} - x^{5}))

What is the tangent of f(x)=x^3-3/2 x^2+x,\at x=1 ?
The tangent of f(x)=x^3-3/2 x^2+x,\at x=1 is y=x-1/2