What is the tangent of f(x)=(x^3-7^x+3),\at x=1 ?

The tangent of f(x)=(x^3-7^x+3),\at x=1 is y=(3-7ln(7))x-6+7ln(7)

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tangent of f(x)=(x^3-7^x+3),\at x=1

tangent of

f (x) = (x^{3} - 7^{x} + 3), at x = 1

y = (3 - 7 ln (7)) x - 6 + 7 ln (7)

Solution steps

Find the tangent point:

(1, - 3)

Find the slope of

f (x) = (x^{3} - 7^{x} + 3) : \frac{df ( x )}{d x} = 3 x^{2} - ln (7) \cdot 7^{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 = 3 - 7 ln (7)

Find the line with slope m=

3 - 7 ln (7)

and passing through

(1, - 3) : y = (3 - 7 ln (7)) x - 6 + 7 ln (7)

y = (3 - 7 ln (7)) x - 6 + 7 ln (7)

y^{^{''}} - 9 y^{^{'}} = 7 t

\int 5 cot^{5} (θ) sin^{4} (θ) d θ

\frac{\partial}{\partial x} (lo g_{10} (x + 1))

x \to 4 lim (\frac{5 x}{x - 4})

\frac{d}{d x} (c x e^{2 x})

What is the tangent of f(x)=(x^3-7^x+3),\at x=1 ?
The tangent of f(x)=(x^3-7^x+3),\at x=1 is y=(3-7ln(7))x-6+7ln(7)