The function is given to be:
[tex]f(x)=x^{3\text{/2}}+x^{1\text{/2}}[/tex]
at (4, 10).
Step 1: Differentiate the function
[tex]\begin{gathered} f^{\prime}(x)=\frac{3}{2}x^{\frac{3}{2}-1}+\frac{1}{2}x^{\frac{1}{2}-1} \\ f^{\prime}(x)=\frac{3}{2}x^{\frac{1}{2}}+\frac{x^{-\frac{1}{2}}}{2} \\ f^{\prime}(x)=\frac{3x^{\frac{1}{2}}}{2}+\frac{1}{2x^{\frac{1}{2}}} \end{gathered}[/tex]
Step 2: Substitute for x = 4 into f'(x)
[tex]\begin{gathered} f^{\prime}(4)=\frac{3(4)^{\frac{1}{2}}}{2}+\frac{1}{2(4)^{\frac{1}{2}}} \\ f^{\prime}(4)=\frac{3\times2}{2}+\frac{1}{2\times2}=3+\frac{1}{4}=\frac{13}{4} \\ \therefore \\ f^{\prime}(4)=\frac{13}{4} \end{gathered}[/tex]
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
[tex]y-y=m(x-x_1)[/tex]
Therefore, we have:
[tex]\begin{gathered} y-10=\frac{13}{4}(x-4) \\ y-10=\frac{13}{4}x-13 \\ y=\frac{13}{4}x-13+10 \\ y=\frac{13}{4}x-3 \end{gathered}[/tex]
ANSWER
The equation of the tangent at the point is:
[tex]y=\frac{13}{4}x-3[/tex]
