What is the tangent of f(x)=x^2+x,\at a=5 ?

The tangent of f(x)=x^2+x,\at a=5 is y=(2a+1)x-a^2

tangent of f(x)=x^2+x,\at a=5

tangent of

f (x) = x^{2} + x, at a = 5

y = (2 a + 1) x - a^{2}

Solution steps

Compute the tangent line to the general point

x = a

Find the tangent point:

(a, a^{2} + a)

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 2 a + 1

Find the line with slope m=

2 a + 1

and passing through

(a, a^{2} + a) : y = (2 a + 1) x - a^{2}

y = (2 a + 1) x - a^{2}

\int \frac{13}{x ( x ^{2} + 4 ) ^{2}} d x

\int \frac{16}{( t ^{2} + 1 ) ^{\frac{3}{2}}} d t

d er i v a t i v e - 4 x + \frac{1}{x} - 3

\frac{\partial}{\partial x} (x^{2} + 2 y + 2 z - 12)

in v erse l a pl a ce \frac{10}{( s + 1 ) ^{2} ( s + 3 )}

What is the tangent of f(x)=x^2+x,\at a=5 ?
The tangent of f(x)=x^2+x,\at a=5 is y=(2a+1)x-a^2