What is the tangent of f(x)=x^3-5x^2+7,\at x=1 ?

The tangent of f(x)=x^3-5x^2+7,\at x=1 is y=-7x+10

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tangent of f(x)=x^3-5x^2+7,\at x=1

tangent of

f (x) = x^{3} - 5 x^{2} + 7, at x = 1

y = - 7 x + 10

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

f (x) = x^{3} - 5 x^{2} + 7 : \frac{df ( x )}{d x} = 3 x^{2} - 10 x

Compute the slope of

f (x) = x^{3} - 5 x^{2} + 7

x = 1 : m = - 7

Find the line with slope m=

- 7

and passing through

(1, 3) : y = - 7 x + 10

y = - 7 x + 10

\int \frac{x + 5}{x ^{2} + 2 x - 15} d x

\int \frac{1}{x ( ln ( x ^{2} ) ) ^{3}} d x

x \to 5 + lim (x^{2} - 25)

\int 0.6 x^{2} d x

x \to + 0 lim (ln (x))

What is the tangent of f(x)=x^3-5x^2+7,\at x=1 ?
The tangent of f(x)=x^3-5x^2+7,\at x=1 is y=-7x+10