What is the tangent of f(x)=7x^2,\at x=4 ?

The tangent of f(x)=7x^2,\at x=4 is y=56x-112

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tangent of f(x)=7x^2,\at x=4

tangent of

f (x) = 7 x^{2}, at x = 4

y = 56 x - 112

Solution steps

Find the tangent point:

(4, 112)

Find the slope of

f (x) = 7 x^{2} : \frac{df ( x )}{d x} = 14 x

Compute the slope of

f (x) = 7 x^{2}

x = 4 : m = 56

Find the line with slope m=

56

and passing through

(4, 112) : y = 56 x - 112

y = 56 x - 112

\frac{d}{d x} (\frac{x + 3}{x - 1})

e^{y} \cdot (\frac{d y}{d x} + 1) = 1

y^{^{''}} + 3 y^{^{'}} + 10 y = 0

(- 6, - 6), (- 5, 0)

\int 16 cos^{2} (x + 3) sin^{2} (x + 3) d x

What is the tangent of f(x)=7x^2,\at x=4 ?
The tangent of f(x)=7x^2,\at x=4 is y=56x-112