Answer to Question #263705 in Calculus for Alunsina

Question

In numbers 1-4,

(a) find the critical numbers of the given functions

(b) using either the First or Second Derivative Test, classify whether thd function has a relative maximum or minimum at these critical numbers

(c) find the coordinates of the exterum points

(d) find the points of jnflection

(e) find the x and y - intercepts

(f) summarize the points in a table. include additional points necessary for the graph

(g) sketch the polunomial curves in graphic papers

1. f(x)=x³+x²-x+2

2. f(x)=(x⁴/4)-(x³/3)-2x²+4x+3

3. f(x)=x⁴-8x²+9

4. f(x)=x³+6x²+9x+3

Answer 1

1(a) Critical points of $f(x)=x³+x²-x+2$

steps

$find\ f'(x)=3x^2+2x-1$

find X

$Critical \ points\ x=-1, x=\frac{1}{3}$

1(b)$f(x)=x³+x²-x+2$

$f'(x)=3x^2+2x-1$

$f'(x)\gt 0:x\lt -1 \ or \ x\gt\ \frac{1}{3}$

$f'(x)\lt0:-1\lt x\lt \frac{1}{3}$

1(c) Coordinates of extrema points $of \ f(x)=x³+x²-x+2$

Local maxima at $x=-1$

$coordinates \ = (-1,3)$

local minima at $x=\frac{1}{3}$

$=(\frac{1}{3},\frac{49}{27})$

1(d)Points of inflection

"f"(x)=6x+2\\\\find \\ where \\ f"(x)=0 \\ or \\ undefined: \\ x=-\\frac{1}{3}"

$plug \ x=-\frac{1}{3} \ into \ f(x)=x³+x²-x+2$

$=(-\frac{1}{3},\frac{65}{27} )$

1(e)

$for \ x \ intercepts \ replace\ y\ with\ 0\\ for \ y \ intercepts \ replace\ x\ with\ 0$

$\mathrm{X\:Intercepts}:\:\left(-2,\:0\right),\: \\ \mathrm{Y\:Intercepts}:\:\left(0,\:2\right)$

1(f) plot the points $for f(x)=x³+x²-x+2$

1(g) Graph $f(x)=x³+x²-x+2$

2(a) $critical \ points\ f\left(x\right)=\left(\frac{x^4}{4}\right)-\left(\frac{x^3}{3}\right)-2x²+4x+3$

$find\ f'(x)=4x^3+3x^2-4x+4\\solve\ for\ x$

$x=-2\\x=1\\x=2$

2(b) first Derivative test for maxima and minimum

$f'(x) = 4x^3-3x^2-4x+4$

$f'(x)\gt0: -2\lt x\lt1\ or \ x\gt2$

$f'(x)\lt0: x\lt-2\ or \ 1\lt x\lt2$

2(c) Coordinates of extrema points

$local\ maxima = (1, \frac{59}{12})\\local\ minima=(-2,-\frac{19}{3} )\\local minima =(2,\frac{13}{3})$

2(d) points of inflection of $f\left(x\right)=\left(\frac{x^4}{4}\right)-\left(\frac{x^3}{3}\right)-2x²+4x+3$

"f"(x)=3x^2-2x-4\\\\find \\ where \\ f"(x)=0 \\ or \\ undefined: \\ x=\\frac{1-\\sqrt{13}}{3}, x=\\frac{1+\\sqrt{13}}{3}"

$plug \ x=x=\frac{1+\sqrt{13}}{3} \ and \ x=\frac{1-\sqrt{13}}{3}\ into \ f(x)=\frac{x^4}{4}-\frac{x^3}{3}-2x^2+4x+3$

$=\left(\frac{1-\sqrt{13}}{3},\:-\frac{2\left(61+35\sqrt{13}\right)}{81}+3\right),\:\left(\frac{1+\sqrt{13}}{3},\:\frac{2\left(35\sqrt{13}-61\right)}{81}+3\right)$

2(e)

$for \ x \ intercepts \ replace\ y\ with\ 0\\ for \ y \ intercepts \ replace\ x\ with\ 0$

$X \ intercepts=\left(-0.59732\dots ,\:0\right),\:\left(-2.87680\dots ,\:0\right) \\\:\mathrm{Y\:Intercepts}=\:\left(0,\:3\right)$

2(f) plot the points$f\left(x\right)=\left(\frac{x^4}{4}\right)-\left(\frac{x^3}{3}\right)-2x²+4x+3$

2(g) graph

3(a)Critical points $of \ f(x)=x^4-8x^2+9$

$find\ f'(x)=4x^3-16x\\solve\ for\ x$

$x=-2,\:x=0,\:x=2$

3(b) using first derivative test

$f'(x)= 4x^3-16x\\f'(x)\gt0: -2\lt x\lt0 \ or \ x\gt2\\f'(x)\lt0: x\lt-2 \ or \ 0\lt x\lt2$

3(c) coordinates of extrema points

$local\ maxima = (0, 9)\\local\ minima=(-2,-7 )\\local\ minima =(2,7)$

3(d) points of inflection $of \ f(x)=x^4-8x^2+9$

"f"(x)=12x^2-16\\\\find \\ where \\ f"(x)=0 \\ or \\ undefined: \\ x=-\\frac{2\\sqrt{3}}{3}, x=\\frac{2\\sqrt{3}}{3}"

$plug \ x=-2\ into \ f(x)=x^4-8x^2+9=\frac{1}{9}$

$=\quad \left(-\frac{2\sqrt{3}}{3},\:\frac{1}{9}\right),\left(\frac{2\sqrt{3}}{3},\:\frac{1}{9}\right)$

3(e)

$for \ x \ intercepts \ replace\ y\ with\ 0\\ for \ y \ intercepts \ replace\ x\ with\ 0$

$\mathrm{X\:Intercepts}:\:\left(\sqrt{4+\sqrt{7}},\:0\right),\:\left(-\sqrt{4+\sqrt{7}},\:0\right),\:\left(\sqrt{4-\sqrt{7}},\:0\right),\:\left(-\sqrt{4-\sqrt{7}},\:0\right),\\ \:\mathrm{ Y\:Intercepts}:\:\left(0,\:9\right)$

3(f) Plot the points $of \ f(x)=x^4-8x^2+9$

3(g) graph

4(a) critical points of f(x)=x³+6x²+9x+3

$find\ f'(x)=3x^2+6x+9\\solve\ for\ x$

$x=-3,\:x=-1$

4(b) $f'(x)=3x^2+12x^2+9$

$f'(x)\gt0 : x\lt-3\ or\ x\gt-1\\f'(x)\lt0: x\lt x\lt-1$

4(c) coordinates of Extrema points of f(x)=x³+6x²+9x+3

$local\ minima = (-1, -1)\\local\ maxima=(-3,-3)$

4(d) Points of inflection of f(x)=x³+6x²+9x+3

"f"(x)=6x+12\\\\find \\ where \\ f"(x)=0 \\ or \\ undefined: \\ x=-2"

$plug \ x=-2\ into \ f(x)=x³+6x²+9x+3=1$

$=\left(-2,\:1\right)$

4(e)

$for \ x \ intercepts \ replace\ y\ with\ 0\\ for \ y \ intercepts \ replace\ x\ with\ 0$

$\mathrm{X\:Intercepts}:\:\left(-0.46791\dots ,\:0\right),\:\left(-1.65270\dots ,\:0\right),\:\left(-3.87938\dots ,\:0\right),\:\\ \mathrm{Y\:Intercepts}:\:\left(0,\:3\right)$

4(f) plot the points of f(x)=x³+6x²+9x+3

4(g) Graph f(x)=x³+6x²+9x+3