What is the tangent of f(x)=x^{1/2}(x^2-x+3),\at x=1 ?

The tangent of f(x)=x^{1/2}(x^2-x+3),\at x=1 is y= 5/2 x+1/2

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tangent of f(x)=x^{1/2}(x^2-x+3),\at x=1

tangent of

f (x) = x^{\frac{1}{2}} (x^{2} - x + 3), at x = 1

y = \frac{5}{2} x + \frac{1}{2}

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

f (x) = x^{\frac{1}{2}} (x^{2} - x + 3) : \frac{df ( x )}{d x} = \frac{5 x ^{2} - 3 x + 3}{2 x ^{\frac{1}{2}}}

Compute the slope of

f (x) = x^{\frac{1}{2}} (x^{2} - x + 3)

x = 1 : m = \frac{5}{2}

Find the line with slope m=

\frac{5}{2}

and passing through

(1, 3) : y = \frac{5}{2} x + \frac{1}{2}

y = \frac{5}{2} x + \frac{1}{2}

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What is the tangent of f(x)=x^{1/2}(x^2-x+3),\at x=1 ?
The tangent of f(x)=x^{1/2}(x^2-x+3),\at x=1 is y= 5/2 x+1/2