20. กำหนดให้ $$ f(x)=2 x^{\frac{5}{2}}-x^{\frac{3}{2}}+x^{\frac{1}{2}} $$ แล้ว $$ \frac{d^{2} y}{d x^{2}} $$ เมื่อ $$ x=1 $$ ตรงกับข้อใด $$ \begin{array}{ll} ext { 1. } 5 x^{\frac{3}{2}}-\frac{3}{2} \sqrt{x} & ext { 2. } 5 x^{\frac{3}{2}}-\frac{3}{2} \sqrt{x}+\frac{1}{2 \sqrt{x}} \ ext { 3. } \frac{11}{2} & ext { 4. } \frac{13}{2} \ ext { 5. } 4 & \end{array} $$

Question

20. กำหนดให้ $$ f(x)=2 x^{\frac{5}{2}}-x^{\frac{3}{2}}+x^{\frac{1}{2}} $$ แล้ว $$ \frac{d^{2} y}{d x^{2}} $$ เมื่อ $$ x=1 $$ ตรงกับข้อใด $$ \begin{array}{ll}	ext { 1. } 5 x^{\frac{3}{2}}-\frac{3}{2} \sqrt{x} & 	ext { 2. } 5 x^{\frac{3}{2}}-\frac{3}{2} \sqrt{x}+\frac{1}{2 \sqrt{x}} \ 	ext { 3. } \frac{11}{2} & 	ext { 4. } \frac{13}{2} \ 	ext { 5. } 4 & \end{array} $$

UpStudy AI · Accepted Answer

Alright, I need to find the second derivative of the function $$ f(x) = 2x^{\frac{5}{2}} - x^{\frac{3}{2}} + x^{\frac{1}{2}} $$ at $$ x = 1 $$. Let's break this down step by step.

First, I'll find the first derivative $$ f'(x) $$. To do that, I'll differentiate each term with respect to $$ x $$:

1. The derivative of $$ 2x^{\frac{5}{2}} $$ is $$ 2 \times \frac{5}{2}x^{\frac{3}{2}} = 5x^{\frac{3}{2}} $$.
2. The derivative of $$ -x^{\frac{3}{2}} $$ is $$ -\frac{3}{2}x^{\frac{1}{2}} $$.
3. The derivative of $$ x^{\frac{1}{2}} $$ is $$ \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}} $$.

So, the first derivative $$ f'(x) $$ is:
$$ f'(x) = 5x^{\frac{3}{2}} - \frac{3}{2}x^{\frac{1}{2}} + \frac{1}{2\sqrt{x}} $$

Next, I'll find the second derivative $$ f''(x) $$ by differentiating $$ f'(x) $$:

1. The derivative of $$ 5x^{\frac{3}{2}} $$ is $$ 5 \times \frac{3}{2}x^{\frac{1}{2}} = \frac{15}{2}x^{\frac{1}{2}} $$.
2. The derivative of $$ -\frac{3}{2}x^{\frac{1}{2}} $$ is $$ -\frac{3}{2} \times \frac{1}{2}x^{-\frac{1}{2}} = -\frac{3}{4}x^{-\frac{1}{2}} = -\frac{3}{4\sqrt{x}} $$.
3. The derivative of $$ \frac{1}{2\sqrt{x}} $$ is $$ \frac{1}{2} \times \left( -\frac{1}{2}x^{-\frac{3}{2}} \right) = -\frac{1}{4}x^{-\frac{3}{2}} = -\frac{1}{4x^{\frac{3}{2}}} $$.

So, the second derivative $$ f''(x) $$ is:
$$ f''(x) = \frac{15}{2}x^{\frac{1}{2}} - \frac{3}{4}x^{-\frac{1}{2}} - \frac{1}{4x^{\frac{3}{2}}} $$

Now, I need to evaluate $$ f''(x) $$ at $$ x = 1 $$:
$$ f''(1) = \frac{15}{2}(1)^{\frac{1}{2}} - \frac{3}{4}(1)^{-\frac{1}{2}} - \frac{1}{4(1)^{\frac{3}{2}}} $$
$$ f''(1) = \frac{15}{2} \times 1 - \frac{3}{4} \times 1 - \frac{1}{4} \times 1 $$
$$ f''(1) = \frac{15}{2} - \frac{3}{4} - \frac{1}{4} $$

To combine these terms, I'll convert them to have a common denominator of 4:
$$ \frac{15}{2} = \frac{30}{4} $$
$$ \frac{30}{4} - \frac{3}{4} - \frac{1}{4} = \frac{30 - 3 - 1}{4} = \frac{26}{4} = \frac{13}{2} $$

Therefore, the value of $$ \frac{d^{2} y}{d x^{2}} $$ at $$ x = 1 $$ is $$ \frac{13}{2} $$.

4

Real Tutor Solution

Answer

Solution

The Deep Dive

Related Questions

Latest Calculus Questions