What is the tangent of f(x)=x^2+x,\at x=1,x=4 ?

The tangent of f(x)=x^2+x,\at x=1,x=4 is y=3x-1

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tangent of f(x)=x^2+x,\at x=1,x=4

tangent of

f (x) = x^{2} + x, at x = 1, x = 4

y = 3 x - 1

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

f (x) = x^{2} + x : \frac{df ( x )}{d x} = 2 x + 1

Compute the slope of

f (x) = x^{2} + x

x = 1 : m = 3

Find the line with slope m=

3

and passing through

(1, 2) : y = 3 x - 1

y = 3 x - 1

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

\int_{- 2}^{0} x^{2} e^{x^{3} + 8} d x

\int \frac{2}{1 + cos ( 2 x )} d x

\int (x - 1) (x + 1) d x

\int x^{3} (2 x^{4} + 4)^{2} d x

What is the tangent of f(x)=x^2+x,\at x=1,x=4 ?
The tangent of f(x)=x^2+x,\at x=1,x=4 is y=3x-1