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

The tangent of f(x)=2x^2+5,\at x=1 is y=4x+3

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tangent of f(x)=2x^2+5,\at x=1

tangent of

f (x) = 2 x^{2} + 5, at x = 1

y = 4 x + 3

Solution steps

Find the tangent point:

(1, 7)

Find the slope of

f (x) = 2 x^{2} + 5 : \frac{df ( x )}{d x} = 4 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 = 4

Find the line with slope m=

4

and passing through

(1, 7) : y = 4 x + 3

y = 4 x + 3

\frac{d}{d x} ((1 + 3 x^{4})^{5})

\int_{0}^{4} (16 - x^{2})^{2} d x

y = (2 x + 3)^{2}

x \to 0 lim (e^{x} ln (x + 1))

\int 4 sin^{3} (x) cos (x) d x

What is the tangent of f(x)=2x^2+5,\at x=1 ?
The tangent of f(x)=2x^2+5,\at x=1 is y=4x+3