What is the tangent of f(x)=41x-16x^2,\at x=1 ?

The tangent of f(x)=41x-16x^2,\at x=1 is y=9x+16

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tangent of f(x)=41x-16x^2,\at x=1

tangent of

f (x) = 41 x - 16 x^{2}, at x = 1

y = 9 x + 16

Solution steps

Find the tangent point:

(1, 25)

Find the slope of

f (x) = 41 x - 16 x^{2} : \frac{df ( x )}{d x} = 41 - 32 x

Compute the slope of

f (x) = 41 x - 16 x^{2}

x = 1 : m = 9

Find the line with slope m=

9

and passing through

(1, 25) : y = 9 x + 16

y = 9 x + 16

\int \frac{1}{1 + sin ( x ) cos ( x )} d x

(1 + x)^{- \frac{6}{5}}

\int t sin (π t) d t

y = (1 - 3 x)^{10}, (0, 1)

\int e^{5 - x} d x

What is the tangent of f(x)=41x-16x^2,\at x=1 ?
The tangent of f(x)=41x-16x^2,\at x=1 is y=9x+16