One hour on earth is four hours on planet S.
1 answer:
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Answer:
Step-by-step explanation:
Given that there is a function of x,
Let us find first and second derivative for f(x)
When f'(x) =0 we have tanx = 1 and hence
a) f'(x) >0 for I and III quadrant
Hence increasing in
and decreasing in
Hence f has a maxima at x = pi/4 and minima at x = 3pi/4
b) Maximum value =
Minimum value =
c)
f"(x) =0 gives tanx =-1
are points of inflection.
concave up in (3pi/4,7pi/4)
and concave down in (0,3pi/4)U(7pi/4,2pi)
Answer: " x = 1 + √5 " or " x = 1 − √5" .
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Given:
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" x² − 2x − <span>4 = 0 " ;
______________________________________________
Solve for "x" by using the "quadratic formula" :
</span>Note: This equation is already written in "quadratic format" ; that is:
" ax² + bx + c = 0 " ; { "a
0" } ;
in which: "a = 1" {the implied coefficient of "1" ;
since "1", multiplied by any value, equals that same value};
"b = -2 " ;
"c = -4 " ;
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The quadratic equation formula:
x = { - b ± √(b² − 4 ac) } / 2a ; {"a0"} ;
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Substitute our known values:
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→ x = { - (-2) ± √[(-2)² − 4(1)(-4)] } / 2(1) ;
→ x = { 2 ± √(4 − 4(-4) } / 2 ;
→ x = { 2 ± √(4 − (-16) } / 2 ;
→ x = { 2 ± √(4 + 16) } / 2 ;
→ x = { 2 ± √(20) } / 2 ;
→ x = { 2 ± √4 √5} / 2 ;
→ x = { 2 ± 2√5} / 2 ;
→ x = 1 ± √5 ;
_______________________________________________________
→ "x = 1 + √5" or "x = 1 − √5" .
_______________________________________________________
______________________________________________________
Given:
______________________________________________________
" x² − 2x − <span>4 = 0 " ;
______________________________________________
Solve for "x" by using the "quadratic formula" :
</span>Note: This equation is already written in "quadratic format" ; that is:
" ax² + bx + c = 0 " ; { "a
in which: "a = 1" {the implied coefficient of "1" ;
since "1", multiplied by any value, equals that same value};
"b = -2 " ;
"c = -4 " ;
_______________________________________________________
The quadratic equation formula:
x = { - b ± √(b² − 4 ac) } / 2a ; {"a
______________________________________________________
Substitute our known values:
______________________________________________________
→ x = { - (-2) ± √[(-2)² − 4(1)(-4)] } / 2(1) ;
→ x = { 2 ± √(4 − 4(-4) } / 2 ;
→ x = { 2 ± √(4 − (-16) } / 2 ;
→ x = { 2 ± √(4 + 16) } / 2 ;
→ x = { 2 ± √(20) } / 2 ;
→ x = { 2 ± √4 √5} / 2 ;
→ x = { 2 ± 2√5} / 2 ;
→ x = 1 ± √5 ;
_______________________________________________________
→ "x = 1 + √5" or "x = 1 − √5" .
_______________________________________________________