for f(x)=4x^(3/2)(5 - x)-x + 1, determine the interval (a,b) where f is concave up. as your answer, please…

Explanation:

Step1: Expand the function

$f(x)=4x^{\frac{3}{2}}(5 - x)-x + 1=20x^{\frac{3}{2}}-4x^{\frac{5}{2}}-x + 1$.

Step2: Find the first - derivative

Using the power rule $(x^n)'=nx^{n - 1}$, we have $f'(x)=20\times\frac{3}{2}x^{\frac{1}{2}}-4\times\frac{5}{2}x^{\frac{3}{2}}-1=30x^{\frac{1}{2}}-10x^{\frac{3}{2}}-1$.

Step3: Find the second - derivative

$f''(x)=30\times\frac{1}{2}x^{-\frac{1}{2}}-10\times\frac{3}{2}x^{\frac{1}{2}}=15x^{-\frac{1}{2}}-15x^{\frac{1}{2}}=\frac{15 - 15x}{x^{\frac{1}{2}}}$.

Step4: Find where $f''(x)>0$

Set $f''(x)>0$, then $\frac{15 - 15x}{x^{\frac{1}{2}}}>0$. Since $x^{\frac{1}{2}}>0$ for the domain $x>0$ (because of the square - root in the original function), we solve $15 - 15x>0$. $15-15x>0$ implies $15>15x$, so $x < 1$. Combining with the domain $x>0$, the interval where $f(x)$ is concave up is $(0,1)$.

Answer:

$1$