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

The tangent of f(x)=sqrt(x^2-x+4),\at x=1 is y= 1/4 x+7/4

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

tangent of

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

y = \frac{1}{4} x + \frac{7}{4}

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

f (x) = x^{2} - x + 4 : \frac{df ( x )}{d x} = \frac{2 x - 1}{2 x ^{2} - x + 4}

Compute the slope of

f (x) = x^{2} - x + 4

x = 1 : m = \frac{1}{4}

Find the line with slope m=

\frac{1}{4}

and passing through

(1, 2) : y = \frac{1}{4} x + \frac{7}{4}

y = \frac{1}{4} x + \frac{7}{4}

f (x) = x^{2} - 4, (4, 12)

\frac{d}{d x} (e^{(5 x)})

\int \frac{4 x + 6}{( x ^{2} + 2 ) ( x + 2 )} d x

(x + y) d x + (x + 2 y) d y = 0

What is the tangent of f(x)=sqrt(x^2-x+4),\at x=1 ?
The tangent of f(x)=sqrt(x^2-x+4),\at x=1 is y= 1/4 x+7/4