Question

9.Evaluate the \\(\\frac{d ^{3}f}{d x^{3}}\\) of \\(f(x)= sin (x) cos (x)\\)
a.\\(\\frac{d ^{3}f}{d x^{3}}=-2\\left(Cos^{2} (x)+sin^{2} (x)\\right)\\)
b.\\(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}}\\)
c.\\(\\frac{d ^{3}f}{d x^{3}}=-4\\left(cos^{2} (x)-sin^{2} (x)\\right)\\)
d.\\(\\frac{d ^{3}f}{d x^{3}}=-4\\left(tan^{2} (x)-cos^{2} (x)\\right)\\)

10.Determine whether the Rolle\'s theorem can be applied to \\(f\\) on the closed interval [a, b] . If can be applied, Find the values of \\(c\\) in open interval (a, b) such that \\(f\'( c) = 0\\), \\(f(x)=\\frac{x^{2}-2x-3}{x+2}, [-1, 3]
a.\\(c=-1\\pm\\sqrt(5)\\)
b.\\(c=-2\\pm\\sqrt(5)\\)
c.\\(c=-2\\pm\\sqrt(5)\\)
d.\\(c=-2\\pm 2\\sqrt(5)\\)

Answer 1

1.

$\frac{}{}$ $\frac{d ^{}f}{d x^{}} \sin (x) \cos (x)=\cos^2x-\sin^2x=\cos2x$

$\frac{d ^{2}}{d x^{2}} \sin (x) \cos (x)=\frac{d ^{}}{d x^{}}\cos2x=-2sin2x$

$\frac{df^3}{dx^3}=\frac{d}{dx}(-2sin2x)=-4cos2x=-4\left(\cos^{2} (x)-\sin^{2} (x)\right)$

answer C

2

b or c

$f(x)=\frac{x^{2}-2x-3}{x+2}$ is continous and differentiable on [-1;3]

$f(-1)=0=f(3)$

Hence, according to Rolle's theorem, there's a $c\in [-1;3]$ such that f'(c)=0

Let's find f'(x)

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

$x^2+4x-1=0\iff x=\frac{-4\pm\sqrt{16+4}}{2}=-2\pm\sqrt{5}$

Of these two roots only $c=-2+\sqrt{5}$ lays inside [-1;3]