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

The tangent of f(x)=x^2+1,\at x=4 is y=8x-15

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tangent of f(x)=x^2+1,\at x=4

tangent of

f (x) = x^{2} + 1, at x = 4

y = 8 x - 15

Solution steps

Find the tangent point:

(4, 17)

Find the slope of

f (x) = x^{2} + 1 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} + 1

x = 4 : m = 8

Find the line with slope m=

8

and passing through

(4, 17) : y = 8 x - 15

y = 8 x - 15

\int \frac{csc ^{2} ( x )}{cot ^{6} ( x )} d x

\int_{2}^{4} e^{t x} \frac{x}{6} d x

\frac{d}{d x} (x^{2} + y^{2} - 2 x y)

\int \frac{1}{t ^{2}} d t

\frac{d}{d x} (\frac{sin ( 3 x )}{1 + x ^{2}})

What is the tangent of f(x)=x^2+1,\at x=4 ?
The tangent of f(x)=x^2+1,\at x=4 is y=8x-15