3/8 mile in 3/4 hour

Mathematics

1 answer:

Elena-2011 [213]3 years ago

5 0

Answer:

1/8

Step-by-step explanation:

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Hard question but would help me out

Alekssandra [29.7K]

Answer:

Critical value f(1)=2.

Minimum at (1,2), function is decreasing for $0$ and increasing for $x>1.$

$\left(3,\dfrac{4}{\sqrt{3}}\right)$ is point of inflection.

When 0<x<3, function is concave upwards and when x>3, , function is concave downwards.

Step-by-step explanation:

1. Find the domain of the function f(x):

$\left\{\begin{array}{l}x\ge 0\\x\neq 0\end{array}\right.\Rightarrow x>0.$

2. Find the derivative f'(x):

$f'(x)=\dfrac{(x+1)'\cdot \sqrt{x}-(x+1)\cdot (\sqrt{x})'}{(\sqrt{x})^2}=\dfrac{\sqrt{x}-\frac{x+1}{2\sqrt{x}}}{x}=\dfrac{2x-x-1}{2x\sqrt{x}}=\dfrac{x-1}{2x^{\frac{3}{2}}}.$

This derivative is equal to 0 at x=1 and is not defined at x=0. Since x=0 is not a point from the domain, the crititcal point is only x=1. The critical value is

$f(1)=\dfrac{1+1}{\sqrt{1}}=2.$

2. For $0$ the derivative f'(x)<0, then the function is decreasing. For $x>1,$ the derivative f'(x)>0, then the function is increasing. This means that point x=1 is point of minimum.

3. Find f''(x):

$f''(x)=\dfrac{(x-1)'\cdot 2x^{\frac{3}{2}}-(x-1)\cdot (2x^{\frac{3}{2}})'}{(2x^{\frac{3}{2}})^2}=$

$=\dfrac{2x^{\frac{3}{2}}-2(x-1)\cdot \frac{3}{2}x^{\frac{1}{2}}}{4x^3}=\dfrac{2x^{\frac{3}{2}}-2\cdot\frac{3}{2}x^{\frac{3}{2}}+ 2\cdot\frac{3}{2}x^{\frac{1}{2}}}{4x^3}=$