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

The tangent of f(x)=2x^2+5x^2-2,\at x=2 is y=28x-30

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

tangent of

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

y = 28 x - 30

Solution steps

Find the tangent point:

(2, 26)

Find the slope of

f (x) = 2 x^{2} + 5 x^{2} - 2 : \frac{df ( x )}{d x} = 14 x

Compute the slope of

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

x = 2 : m = 28

Find the line with slope m=

28

and passing through

(2, 26) : y = 28 x - 30

y = 28 x - 30

y = x^{2} + 6

\int_{- 1}^{1} x^{3} - x d x

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

\int_{0}^{\infty} sin^{2} (x) d x

x \to 5 lim ((3 x^{2} - x - 2) 0)

What is the tangent of f(x)=2x^2+5x^2-2,\at x=2 ?
The tangent of f(x)=2x^2+5x^2-2,\at x=2 is y=28x-30