Answer to Question #94031 in Calculus for Jflows

Question

3.Find the two x-intercept of \\(f(x)=x^{2}-3x+2\\)
a.x=-2, 2
b.x= 1, 2
c.x=1, 1
d.x=1, 3

4.Compute the first thrre derivatives of \\(f(x)=2x^{5}+x^{\\frac{3}{2}}-\\frac{1}{2x}\\)
a.\\(f\'(x)=5x^{4}-\\frac{1}{2}x^{\\frac{1}{2}}+ \\frac{1}{2x^{2}}, 20x^{3}-\\frac{3}{4}x^{-\\frac{1}{2}}- \\frac{1}{x^{1}}, 100x^{2}-\\frac{3}{8}x^{-\\frac{3}{2}}+ \\frac{3}{x^{4}}\\)
b.\\(f\'(x)=10x^{4}-\\frac{3}{2}x^{\\frac{2}{2}}-\\frac{1}{2x^{2}}, 40x^{3}\\frac{3}{4}x^{-\\frac{1}{2}}- \\frac{1}{x^{3}}, 120x^{2}-\\frac{3}{8}x^{-\\frac{1}{2}}+ \\frac{3}{x^{4}}\\)
c.\\(f\'(x)=10x^{3}-\\frac{2}{2}x^{\\frac{1}{2}}+ \\frac{1}{2x^{2}}, 20x^{3}-\\frac{3}{4}x^{-\\frac{1}{2}}- \\frac{1}{x^{3}}, 10x^{2}-\\frac{1}{8}x^{-\\frac{3}{2}}+ \\frac{3}{x^{4}}\\)
d.\\(f\'(x)=10x^{4}-\\frac{3}{2}x^{\\frac{1}{2}}+ \\frac{1}{2x^{2}}, 40x^{3}-\\frac{3}{4}x^{-\\frac{1}{2}}- \\frac{1}{x^{3}}, 120x^{2}-\\frac{3}{8}x^{-\\frac{3}{2}}+ \\frac{3}{x^{4}}\\)

Answer 1

3. Find the two x-intercept of $f(x)=x^{2}-3x+2$.

Solve the equation $x^{2}-3x+2=0$

$D=b^2-4ac=3^2-4\cdot 1\cdot 2=9-8=1;\\ \sqrt{D}=1;\\ x_1=\frac{-b+\sqrt{D}}{2a}=\frac{3+1}{2}=2;\\ x_2=\frac{-b-\sqrt{D}}{2a}=\frac{3-1}{2}=1.$

Answer: b. $x$ = 1, 2 

4. Compute the first three derivatives of $f(x)=2x^{5}+x^{\frac{3}{2}}-\frac{1}{2x}$

Note that $(x^n)'=nx^{n-1}$ and $f(x)=2x^{5}+x^{\frac{3}{2}}-\frac{1}{2x}=2x^{5}+x^{\frac{3}{2}}-\frac{1}{2}x^{-1}$ .

We get

$f'(x)=2\cdot 5x^4+\frac{3}{2}x^{\frac{1}{2}}-\frac{1}{2}\cdot (-1)x^{-2}=\\ = 10x^4+\frac{3}{2}x^{\frac{1}{2}}+\frac{1}{2}x^{-2}=\\ =10x^4+\frac{3}{2}x^{\frac{1}{2}}+\frac{1}{2x^2};$

$f''(x)=10\cdot 4x^3+\frac{3}{2}\cdot\frac{1}{2}x^{-\frac{1}{2}}+\frac{1}{2}\cdot (-2)x^{-3}=\\ = 40x^3+\frac{3}{4}x^{-\frac{1}{2}}-\frac{1}{x^3};$

$f'''(x)=40\cdot 3x^2+\frac{3}{4}\cdot(-\frac{1}{2})x^{-\frac{3}{2}}-(-3)x^{-4}=\\ =120x^2-\frac{3}{8}x^{-\frac{3}{2}}+\frac{3}{x^4}.$

Answer: d. $10x^{4}-\frac{3}{2}x^{\frac{1}{2}}+ \frac{1}{2x^{2}}, 40x^{3}-\frac{3}{4}x^{-\frac{1}{2}}- \frac{1}{x^{3}}, 120x^{2}-\frac{3}{8}x^{-\frac{3}{2}}+ \frac{3}{x^{4}}$