Question

Find an equation of the tangent line at the indicated point.

Answer 1

The function is given to be:

`f(x)=x^{3\text{/2}}+x^{1\text{/2}}`

at (4, 10).

Step 1: Differentiate the function

`\begin{gathered} f^(\prime)(x)=(3)/(2)x^{(3)/(2)-1}+(1)/(2)x^{(1)/(2)-1} \\ f^(\prime)(x)=(3)/(2)x^{(1)/(2)}+\frac{x^{-(1)/(2)}}{2} \\ f^(\prime)(x)=\frac{3x^{(1)/(2)}}{2}+\frac{1}{2x^{(1)/(2)}} \end{gathered}`

Step 2: Substitute for x = 4 into f'(x)

`\begin{gathered} f^(\prime)(4)=\frac{3(4)^{(1)/(2)}}{2}+\frac{1}{2(4)^{(1)/(2)}} \\ f^(\prime)(4)=(3*2)/(2)+(1)/(2*2)=3+(1)/(4)=(13)/(4) \\ \therefore \\ f^(\prime)(4)=(13)/(4) \end{gathered}`

Step 3: Find a line with the slope 13/4 passing through (4, 10) using the point-slope form of the equation of a straight line given to be

`y-y=m(x-x_1)`

Therefore, we have:

`\begin{gathered} y-10=(13)/(4)(x-4) \\ y-10=(13)/(4)x-13 \\ y=(13)/(4)x-13+10 \\ y=(13)/(4)x-3 \end{gathered}`

ANSWER

The equation of the tangent at the point is:

`y=(13)/(4)x-3`