What is the tangent of f(x)=5x-9x^2,\at x=1 ?

The tangent of f(x)=5x-9x^2,\at x=1 is y=-13x+9

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tangent of f(x)=5x-9x^2,\at x=1

tangent of

f (x) = 5 x - 9 x^{2}, at x = 1

y = - 13 x + 9

Solution steps

Find the tangent point:

(1, - 4)

Find the slope of

f (x) = 5 x - 9 x^{2} : \frac{df ( x )}{d x} = 5 - 18 x

Compute the slope of

f (x) = 5 x - 9 x^{2}

x = 1 : m = - 13

Find the line with slope m=

- 13

and passing through

(1, - 4) : y = - 13 x + 9

y = - 13 x + 9

\frac{d}{d x} (- 8 cos (x))

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\int \frac{8 + x}{64 - x ^{2}} d x

\frac{d}{d x} (2 x (cos (x^{2})))

\frac{d}{d x} (x^{3} - 12 x + 2)

What is the tangent of f(x)=5x-9x^2,\at x=1 ?
The tangent of f(x)=5x-9x^2,\at x=1 is y=-13x+9