What is the tangent of f(x)=(x^2+7)(5-x),\at x=0 ?

The tangent of f(x)=(x^2+7)(5-x),\at x=0 is y=-7x+35

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tangent of f(x)=(x^2+7)(5-x),\at x=0

tangent of

f (x) = (x^{2} + 7) (5 - x), at x = 0

y = - 7 x + 35

Solution steps

Find the tangent point:

(0, 35)

Find the slope of

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

Compute the slope of

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

x = 0 : m = - 7

Find the line with slope m=

- 7

and passing through

(0, 35) : y = - 7 x + 35

y = - 7 x + 35

\frac{d}{d x} (x^{4} + (y - 2)^{3})

\int \frac{( x ^{2} - 2 x + 1 ) ^{\frac{1}{5}}}{1 - x} d x

\int x sin^{2} (9 x) d x

2 x^{\frac{3}{2}}

\int sin^{3} (5 x) cos^{- 2} (5 x) d x

What is the tangent of f(x)=(x^2+7)(5-x),\at x=0 ?
The tangent of f(x)=(x^2+7)(5-x),\at x=0 is y=-7x+35