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

The tangent of f(x)=x^2,\at x=7 is y=14x-49

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

tangent of

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

y = 14 x - 49

Solution steps

Find the tangent point:

(7, 49)

Find the slope of

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

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 = 14

Find the line with slope m=

14

and passing through

(7, 49) : y = 14 x - 49

y = 14 x - 49

y^{^{''}} + 3 y^{^{'}} + 9 y = 0

h \to 6 lim (\frac{4 h}{h - 6})

\frac{d}{d x} (- tan (9 x))

x \to 0 lim ([(2 x + 1) (x - 3)])

y = 2 x + \frac{5}{x}

What is the tangent of f(x)=x^2,\at x=7 ?
The tangent of f(x)=x^2,\at x=7 is y=14x-49