What is the tangent of f(x)=6x^4+7x-8,(1,5) ?

The tangent of f(x)=6x^4+7x-8,(1,5) is y=31x-26

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tangent of f(x)=6x^4+7x-8,(1,5)

tangent of

f (x) = 6 x^{4} + 7 x - 8, (1, 5)

y = 31 x - 26

Solution steps

Find the slope of

f (x) = 6 x^{4} + 7 x - 8 : \frac{df ( x )}{d x} = 24 x^{3} + 7

Compute the slope of

f (x) = 6 x^{4} + 7 x - 8

(1, 5) : m = 31

Find the line with slope m=

31

and passing through

(1, 5) : y = 31 x - 26

y = 31 x - 26

\int_{0}^{\frac{1}{4}} 2 arcsin (4 x) d x

\int \frac{v ( x )}{cos ( x )}

x \to 2 lim (x^{5} - 32 x)

2 (x^{2} + y^{2}) d x + (7 x y) d y = 0

\int_{0}^{2} \frac{x ^{2}}{x ^{2} + 4} d x

What is the tangent of f(x)=6x^4+7x-8,(1,5) ?
The tangent of f(x)=6x^4+7x-8,(1,5) is y=31x-26