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

The tangent of f(x)=x^2-sqrt(x),\at x=4 is y= 31/4 x-17

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

tangent of

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

y = \frac{31}{4} x - 17

Solution steps

Find the tangent point:

(4, 14)

Find the slope of

f (x) = x^{2} - x : \frac{df ( x )}{d x} = 2 x - \frac{1}{2 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 = \frac{31}{4}

Find the line with slope m=

\frac{31}{4}

and passing through

(4, 14) : y = \frac{31}{4} x - 17

y = \frac{31}{4} x - 17

What is the tangent of f(x)=x^2-sqrt(x),\at x=4 ?
The tangent of f(x)=x^2-sqrt(x),\at x=4 is y= 31/4 x-17