What is the tangent of f(x)=3x^3-x^2+5,(2,25) ?

The tangent of f(x)=3x^3-x^2+5,(2,25) is y=32x-39

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tangent of f(x)=3x^3-x^2+5,(2,25)

tangent of

f (x) = 3 x^{3} - x^{2} + 5, (2, 25)

y = 32 x - 39

Solution steps

Find the slope of

f (x) = 3 x^{3} - x^{2} + 5 : \frac{df ( x )}{d x} = 9 x^{2} - 2 x

Compute the slope of

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

(2, 25) : m = 32

Find the line with slope m=

32

and passing through

(2, 25) : y = 32 x - 39

y = 32 x - 39

y^{^{'}} + 4 x y = x^{3} e^{x^{2}}

d x = 4 (x^{2} + 1) d t

\frac{1}{( s + 5 )}

\frac{1}{5 s - 2}

\int \frac{x ^{2}}{2 x} d x

What is the tangent of f(x)=3x^3-x^2+5,(2,25) ?
The tangent of f(x)=3x^3-x^2+5,(2,25) is y=32x-39