What is the tangent of f(x)=8x-x^2,\at x=1 ?

The tangent of f(x)=8x-x^2,\at x=1 is y=6x+1

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tangent of f(x)=8x-x^2,\at x=1

tangent of

f (x) = 8 x - x^{2}, at x = 1

y = 6 x + 1

Solution steps

Find the tangent point:

(1, 7)

Find the slope of

f (x) = 8 x - x^{2} : \frac{df ( x )}{d x} = 8 - 2 x

Compute the slope of

f (x) = 8 x - x^{2}

x = 1 : m = 6

Find the line with slope m=

6

and passing through

(1, 7) : y = 6 x + 1

y = 6 x + 1

\int (2 + 3 x^{2} - 8 x^{3}) d x

y^{^{''}} + 3 y^{^{'}} + 2 y = 15 + e x

\int x - x^{3} d x

\frac{d}{d x} (a c e^{- e^{c (b - x)} + c (b - x)})

\int tan (x) sec^{6} (x) d x

What is the tangent of f(x)=8x-x^2,\at x=1 ?
The tangent of f(x)=8x-x^2,\at x=1 is y=6x+1