1(a) Critical points of f ( x ) = x 3 + x 2 − x + 2 f(x)=x³+x²-x+2 f ( x ) = x 3 + x 2 − x + 2
steps
f i n d f ′ ( x ) = 3 x 2 + 2 x − 1 find\ f'(x)=3x^2+2x-1 f in d f ′ ( x ) = 3 x 2 + 2 x − 1
find X
C r i t i c a l p o i n t s x = − 1 , x = 1 3 Critical \ points\ x=-1, x=\frac{1}{3} C r i t i c a l p o in t s x = − 1 , x = 3 1
1(b)f ( x ) = x 3 + x 2 − x + 2 f(x)=x³+x²-x+2 f ( x ) = x 3 + x 2 − x + 2
f ′ ( x ) = 3 x 2 + 2 x − 1 f'(x)=3x^2+2x-1 f ′ ( x ) = 3 x 2 + 2 x − 1
f ′ ( x ) > 0 : x < − 1 o r x > 1 3 f'(x)\gt 0:x\lt -1 \ or \ x\gt\ \frac{1}{3} f ′ ( x ) > 0 : x < − 1 or x > 3 1
f ′ ( x ) < 0 : − 1 < x < 1 3 f'(x)\lt0:-1\lt x\lt \frac{1}{3} f ′ ( x ) < 0 : − 1 < x < 3 1
1(c) Coordinates of extrema points o f f ( x ) = x 3 + x 2 − x + 2 of \ f(x)=x³+x²-x+2 o f f ( x ) = x 3 + x 2 − x + 2
Local maxima at x = − 1 x=-1 x = − 1
c o o r d i n a t e s = ( − 1 , 3 ) coordinates \ = (-1,3) coor d ina t es = ( − 1 , 3 )
local minima at x = 1 3 x=\frac{1}{3} x = 3 1
= ( 1 3 , 49 27 ) =(\frac{1}{3},\frac{49}{27}) = ( 3 1 , 27 49 )
1(d)Points of inflection
f " ( x ) = 6 x + 2 f i n d w h e r e f " ( x ) = 0 o r u n d e f i n e d : x = − 1 3 f"(x)=6x+2\\find \ where \ f"(x)=0 \ or \ undefined: \ x=-\frac{1}{3} f " ( x ) = 6 x + 2 f in d w h ere f " ( x ) = 0 or u n d e f in e d : x = − 3 1
p l u g x = − 1 3 i n t o f ( x ) = x 3 + x 2 − x + 2 plug \ x=-\frac{1}{3} \ into \ f(x)=x³+x²-x+2 pl ug x = − 3 1 in t o f ( x ) = x 3 + x 2 − x + 2
= ( − 1 3 , 65 27 ) =(-\frac{1}{3},\frac{65}{27} ) = ( − 3 1 , 27 65 )
1(e)
f o r x i n t e r c e p t s r e p l a c e y w i t h 0 f o r y i n t e r c e p t s r e p l a c e x w i t h 0 for \ x \ intercepts \ replace\ y\ with\ 0\\ for \ y \ intercepts \ replace\ x\ with\ 0 f or x in t erce pt s re pl a ce y w i t h 0 f or y in t erce pt s re pl a ce x w i t h 0
X I n t e r c e p t s : ( − 2 , 0 ) , Y I n t e r c e p t s : ( 0 , 2 ) \mathrm{X\:Intercepts}:\:\left(-2,\:0\right),\: \\ \mathrm{Y\:Intercepts}:\:\left(0,\:2\right) X Intercepts : ( − 2 , 0 ) , Y Intercepts : ( 0 , 2 )
1(f) plot the points f o r f ( x ) = x 3 + x 2 − x + 2 for f(x)=x³+x²-x+2 f or f ( x ) = x 3 + x 2 − x + 2
1(g) Graph f ( x ) = x 3 + x 2 − x + 2 f(x)=x³+x²-x+2 f ( x ) = x 3 + x 2 − x + 2
2(a) c r i t i c a l p o i n t s f ( x ) = ( x 4 4 ) − ( x 3 3 ) − 2 x 2 + 4 x + 3 critical \ points\ f\left(x\right)=\left(\frac{x^4}{4}\right)-\left(\frac{x^3}{3}\right)-2x²+4x+3 cr i t i c a l p o in t s f ( x ) = ( 4 x 4 ) − ( 3 x 3 ) − 2 x 2 + 4 x + 3
f i n d f ′ ( x ) = 4 x 3 + 3 x 2 − 4 x + 4 s o l v e f o r x find\ f'(x)=4x^3+3x^2-4x+4\\solve\ for\ x f in d f ′ ( x ) = 4 x 3 + 3 x 2 − 4 x + 4 so l v e f or x
x = − 2 x = 1 x = 2 x=-2\\x=1\\x=2 x = − 2 x = 1 x = 2
2(b) first Derivative test for maxima and minimum
f ′ ( x ) = 4 x 3 − 3 x 2 − 4 x + 4 f'(x) = 4x^3-3x^2-4x+4 f ′ ( x ) = 4 x 3 − 3 x 2 − 4 x + 4
f ′ ( x ) > 0 : − 2 < x < 1 o r x > 2 f'(x)\gt0: -2\lt x\lt1\ or \ x\gt2 f ′ ( x ) > 0 : − 2 < x < 1 or x > 2
f ′ ( x ) < 0 : x < − 2 o r 1 < x < 2 f'(x)\lt0: x\lt-2\ or \ 1\lt x\lt2 f ′ ( x ) < 0 : x < − 2 or 1 < x < 2
2(c) Coordinates of extrema points
l o c a l m a x i m a = ( 1 , 59 12 ) l o c a l m i n i m a = ( − 2 , − 19 3 ) l o c a l m i n i m a = ( 2 , 13 3 ) local\ maxima = (1, \frac{59}{12})\\local\ minima=(-2,-\frac{19}{3} )\\local minima =(2,\frac{13}{3}) l oc a l ma x ima = ( 1 , 12 59 ) l oc a l minima = ( − 2 , − 3 19 ) l oc a l minima = ( 2 , 3 13 )
2(d) points of inflection of f ( x ) = ( x 4 4 ) − ( x 3 3 ) − 2 x 2 + 4 x + 3 f\left(x\right)=\left(\frac{x^4}{4}\right)-\left(\frac{x^3}{3}\right)-2x²+4x+3 f ( x ) = ( 4 x 4 ) − ( 3 x 3 ) − 2 x 2 + 4 x + 3
f " ( x ) = 3 x 2 − 2 x − 4 f i n d w h e r e f " ( x ) = 0 o r u n d e f i n e d : x = 1 − 13 3 , x = 1 + 13 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} f " ( x ) = 3 x 2 − 2 x − 4 f in d w h ere f " ( x ) = 0 or u n d e f in e d : x = 3 1 − 13 , x = 3 1 + 13
p l u g x = x = 1 + 13 3 a n d x = 1 − 13 3 i n t o f ( x ) = x 4 4 − x 3 3 − 2 x 2 + 4 x + 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 pl ug x = x = 3 1 + 13 an d x = 3 1 − 13 in t o f ( x ) = 4 x 4 − 3 x 3 − 2 x 2 + 4 x + 3
= ( 1 − 13 3 , − 2 ( 61 + 35 13 ) 81 + 3 ) , ( 1 + 13 3 , 2 ( 35 13 − 61 ) 81 + 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) = ( 3 1 − 13 , − 81 2 ( 61 + 35 13 ) + 3 ) , ( 3 1 + 13 , 81 2 ( 35 13 − 61 ) + 3 )
2(e)
f o r x i n t e r c e p t s r e p l a c e y w i t h 0 f o r y i n t e r c e p t s r e p l a c e x w i t h 0 for \ x \ intercepts \ replace\ y\ with\ 0\\ for \ y \ intercepts \ replace\ x\ with\ 0 f or x in t erce pt s re pl a ce y w i t h 0 f or y in t erce pt s re pl a ce x w i t h 0
X i n t e r c e p t s = ( − 0.59732 … , 0 ) , ( − 2.87680 … , 0 ) Y I n t e r c e p t s = ( 0 , 3 ) X \ intercepts=\left(-0.59732\dots ,\:0\right),\:\left(-2.87680\dots ,\:0\right)
\\\:\mathrm{Y\:Intercepts}=\:\left(0,\:3\right) X in t erce pt s = ( − 0.59732 … , 0 ) , ( − 2.87680 … , 0 ) Y Intercepts = ( 0 , 3 )
2(f) plot the pointsf ( x ) = ( x 4 4 ) − ( x 3 3 ) − 2 x 2 + 4 x + 3 f\left(x\right)=\left(\frac{x^4}{4}\right)-\left(\frac{x^3}{3}\right)-2x²+4x+3 f ( x ) = ( 4 x 4 ) − ( 3 x 3 ) − 2 x 2 + 4 x + 3
2(g) graph
3(a)Critical points o f f ( x ) = x 4 − 8 x 2 + 9 of \ f(x)=x^4-8x^2+9 o f f ( x ) = x 4 − 8 x 2 + 9
f i n d f ′ ( x ) = 4 x 3 − 16 x s o l v e f o r x find\ f'(x)=4x^3-16x\\solve\ for\ x f in d f ′ ( x ) = 4 x 3 − 16 x so l v e f or x
x = − 2 , x = 0 , x = 2 x=-2,\:x=0,\:x=2 x = − 2 , x = 0 , x = 2
3(b) using first derivative test
f ′ ( x ) = 4 x 3 − 16 x f ′ ( x ) > 0 : − 2 < x < 0 o r x > 2 f ′ ( x ) < 0 : x < − 2 o r 0 < x < 2 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 f ′ ( x ) = 4 x 3 − 16 x f ′ ( x ) > 0 : − 2 < x < 0 or x > 2 f ′ ( x ) < 0 : x < − 2 or 0 < x < 2
3(c) coordinates of extrema points
l o c a l m a x i m a = ( 0 , 9 ) l o c a l m i n i m a = ( − 2 , − 7 ) l o c a l m i n i m a = ( 2 , 7 ) local\ maxima = (0, 9)\\local\ minima=(-2,-7 )\\local\ minima =(2,7) l oc a l ma x ima = ( 0 , 9 ) l oc a l minima = ( − 2 , − 7 ) l oc a l minima = ( 2 , 7 )
3(d) points of inflection o f f ( x ) = x 4 − 8 x 2 + 9 of \ f(x)=x^4-8x^2+9 o f f ( x ) = x 4 − 8 x 2 + 9
f " ( x ) = 12 x 2 − 16 f i n d w h e r e f " ( x ) = 0 o r u n d e f i n e d : x = − 2 3 3 , x = 2 3 3 f"(x)=12x^2-16\\find \ where \ f"(x)=0 \ or \ undefined: \ x=-\frac{2\sqrt{3}}{3}, x=\frac{2\sqrt{3}}{3} f " ( x ) = 12 x 2 − 16 f in d w h ere f " ( x ) = 0 or u n d e f in e d : x = − 3 2 3 , x = 3 2 3
p l u g x = − 2 i n t o f ( x ) = x 4 − 8 x 2 + 9 = 1 9 plug \ x=-2\ into \ f(x)=x^4-8x^2+9=\frac{1}{9} pl ug x = − 2 in t o f ( x ) = x 4 − 8 x 2 + 9 = 9 1
= ( − 2 3 3 , 1 9 ) , ( 2 3 3 , 1 9 ) =\quad \left(-\frac{2\sqrt{3}}{3},\:\frac{1}{9}\right),\left(\frac{2\sqrt{3}}{3},\:\frac{1}{9}\right) = ( − 3 2 3 , 9 1 ) , ( 3 2 3 , 9 1 )
3(e)
f o r x i n t e r c e p t s r e p l a c e y w i t h 0 f o r y i n t e r c e p t s r e p l a c e x w i t h 0 for \ x \ intercepts \ replace\ y\ with\ 0\\ for \ y \ intercepts \ replace\ x\ with\ 0 f or x in t erce pt s re pl a ce y w i t h 0 f or y in t erce pt s re pl a ce x w i t h 0
X I n t e r c e p t s : ( 4 + 7 , 0 ) , ( − 4 + 7 , 0 ) , ( 4 − 7 , 0 ) , ( − 4 − 7 , 0 ) , Y I n t e r c e p t s : ( 0 , 9 ) \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) X Intercepts : ( 4 + 7 , 0 ) , ( − 4 + 7 , 0 ) , ( 4 − 7 , 0 ) , ( − 4 − 7 , 0 ) , Y Intercepts : ( 0 , 9 )
3(f) Plot the points o f f ( x ) = x 4 − 8 x 2 + 9 of \ f(x)=x^4-8x^2+9 o f f ( x ) = x 4 − 8 x 2 + 9
3(g) graph
4(a) critical points of f(x)=x³+6x²+9x+3
f i n d f ′ ( x ) = 3 x 2 + 6 x + 9 s o l v e f o r x find\ f'(x)=3x^2+6x+9\\solve\ for\ x f in d f ′ ( x ) = 3 x 2 + 6 x + 9 so l v e f or x
x = − 3 , x = − 1 x=-3,\:x=-1 x = − 3 , x = − 1
4(b) f ′ ( x ) = 3 x 2 + 12 x 2 + 9 f'(x)=3x^2+12x^2+9 f ′ ( x ) = 3 x 2 + 12 x 2 + 9
f ′ ( x ) > 0 : x < − 3 o r x > − 1 f ′ ( x ) < 0 : x < x < − 1 f'(x)\gt0 : x\lt-3\ or\ x\gt-1\\f'(x)\lt0: x\lt x\lt-1 f ′ ( x ) > 0 : x < − 3 or x > − 1 f ′ ( x ) < 0 : x < x < − 1
4(c) coordinates of Extrema points of f(x)=x³+6x²+9x+3
l o c a l m i n i m a = ( − 1 , − 1 ) l o c a l m a x i m a = ( − 3 , − 3 ) local\ minima = (-1, -1)\\local\ maxima=(-3,-3) l oc a l minima = ( − 1 , − 1 ) l oc a l ma x ima = ( − 3 , − 3 )
4(d) Points of inflection of f(x)=x³+6x²+9x+3
f " ( x ) = 6 x + 12 f i n d w h e r e f " ( x ) = 0 o r u n d e f i n e d : x = − 2 f"(x)=6x+12\\find \ where \ f"(x)=0 \ or \ undefined: \ x=-2 f " ( x ) = 6 x + 12 f in d w h ere f " ( x ) = 0 or u n d e f in e d : x = − 2
p l u g x = − 2 i n t o f ( x ) = x 3 + 6 x 2 + 9 x + 3 = 1 plug \ x=-2\ into \ f(x)=x³+6x²+9x+3=1 pl ug x = − 2 in t o f ( x ) = x 3 + 6 x 2 + 9 x + 3 = 1
= ( − 2 , 1 ) =\left(-2,\:1\right) = ( − 2 , 1 )
4(e)
f o r x i n t e r c e p t s r e p l a c e y w i t h 0 f o r y i n t e r c e p t s r e p l a c e x w i t h 0 for \ x \ intercepts \ replace\ y\ with\ 0\\ for \ y \ intercepts \ replace\ x\ with\ 0 f or x in t erce pt s re pl a ce y w i t h 0 f or y in t erce pt s re pl a ce x w i t h 0
X I n t e r c e p t s : ( − 0.46791 … , 0 ) , ( − 1.65270 … , 0 ) , ( − 3.87938 … , 0 ) , Y I n t e r c e p t s : ( 0 , 3 ) \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) X Intercepts : ( − 0.46791 … , 0 ) , ( − 1.65270 … , 0 ) , ( − 3.87938 … , 0 ) , Y Intercepts : ( 0 , 3 )
4(f) plot the points of f(x)=x³+6x²+9x+3
4(g) Graph f(x)=x³+6x²+9x+3
#339914 on Dec 2023
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