What is the tangent of f(x)=15x^3-10x^2+1,(1,6) ?

The tangent of f(x)=15x^3-10x^2+1,(1,6) is y=25x-19

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tangent of f(x)=15x^3-10x^2+1,(1,6)

tangent of

f (x) = 15 x^{3} - 10 x^{2} + 1, (1, 6)

y = 25 x - 19

Solution steps

Find the slope of

f (x) = 15 x^{3} - 10 x^{2} + 1 : \frac{df ( x )}{d x} = 45 x^{2} - 20 x

Compute the slope of

f (x) = 15 x^{3} - 10 x^{2} + 1

(1, 6) : m = 25

Find the line with slope m=

25

and passing through

(1, 6) : y = 25 x - 19

y = 25 x - 19

\int_{b}^{1} 5 ln (x) d x

y^{^{''}} - \frac{y}{4} = 0

\frac{d}{d x} (\frac{4}{16 x ^{2} + 1})

\frac{\partial}{\partial x} (e^{x y})

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

What is the tangent of f(x)=15x^3-10x^2+1,(1,6) ?
The tangent of f(x)=15x^3-10x^2+1,(1,6) is y=25x-19