Line 1 executes n times.Line 2 executes n times.Lines 3 executes one more time than its previous iteration. In other words, it executes 1 time when j = i, then it executes two times when j = 2 all the way until j = n. So you have 1 + 2 + 3 + .. + (n - 2) + (n - 1) + n. This is the sum of the first n integers which equals (n * (n + 1)) / 2 = (n^2 + n) / 2. Dropping the coefficient and keeping only the dominant term you get n ^2.Since the 3 for loops are nested you need to multiply each of the execution times together which is n * n * n^2 = n ^4. This however doesn’t mean the runtime is n ^4 yet as you have to see what the run time is for the other 2 nested loops.Lines 5 executes n^2 times.Lines 6 executes the same way lines 3 does but it runs from 0 .. n ^ 2. So the formula is (n^2 * (n ^ 2 + 1)) / 2 = (n^4 + n ^2) / 2. Dropping constants and keeping only the dominant term you get n ^4.Since you have two groups of nested for loops you need to add the run times together so now you have n^4 + n^4 = 2n^4. Dropping the coefficient you get n ^4. That’s the reasoning.
What is the anti-derivative of (1-x) / (1+x^2)? Please can you explain step by step thanks?We have ∫(1-x)/(1+x²) dxWe have a quotient of functions with a square in the denominator. This looks like a case of trigonometric substitution. Our denominator is 1+x², which would be simplified by the trig identity 1+tan²x=sec²x , so lets's substitutex=tan(u)dx=sec²(u) duPlug in for x and dx:∫(1-tan(u))/(1+tan²(u)) *sec²(u) du∫(1-tan(u))*sec²(u)/(1+tan²(u)) duNote the Pythagorean Identity 1+tan²(u)=sec²(u)This lets us substitute ∫(1-tan(u))*sec²(u)/(sec²(u)) duThe sec²(u) cancels and we're left with∫1-tan(u) duWe know tan(u)=sin(u)/cos(u)∫1-(sin(u)/cos(u)) duLet's integrate that 1 to simplify the antiderivativeu-∫(sin(u)/cos(u)) duNow we have more substitution. Let's letv=cos(u)dv=-sin(u) duPlug in for v and dvu-∫-1/v dvNegatives cancelu+∫1/v dvNow we can integrate directlyu+ln(v)Now we need to get back into terms of x. So we un-substitute v=cos(u) to getu+ln(cos(u))And we defined x=tan(u), so solving for u givesu=arctan(x)Resubsitutearctan(x)+ln(cos(arctan(x)))+CAnd don't forget the constant of integration C
F(x)=k-x^2 y= -6x+1. Find k, such that the line is tangent to the graph of the function. Please explain the process . Thanks?
to be a tangent the curve should have equal slope to the line ..hence the slope of the curve should have a point x where its derivative =- 6 f ' (x) = -2x =-6 x = 3 at x= 3 , the line has coordinate y = (-6)(3) + 1 = -18+1 = -17 the curve -17 = k - (3)^2 -17= k -9 k = -17+9 = -8
Let f(x)= 8x^2 - 5x and g(x)= 7x +9 find the composite f [g(-3)] A. 1212 B. 408 C. 7050 D. 618?
f(x)= 8x² - 5x and g(x)= 7x + 9 fog(x) = f(g(x)) = f(7x + 9) = 8(7x + 9)² - 5(7x + 9) f[g(-3)] = 8(7(-3) + 9)² - 5(7(-3) + 9) = 8(-21 + 9)² - 5(-21 + 9) = 1152 + 60 = 1212
Find and simplify (f(a+h)-f(a))/h f(x)=8x^2-6x+1?
Hello, we have: f(x) = 8x² - 6x + 1 hence: f(a) = 8(a)² - 6(a) + 1 = 8a² - 6a + 1 f(a + h) = 8(a + h)² - 6(a + h) + 1 = 8(a² + 2ah + h²) - 6a - 6h + 1 = 8a² + 16ah + 8h² - 6a - 6h + 1 and therefore: [f(a + h) - f(a)] /h = [(8a² + 16ah + 8h² - 6a - 6h + 1) - (8a² - 6a + 1)] /h = (8a² + 16ah + 8h² - 6a - 6h + 1 - 8a² + 6a - 1) /h = (16ah + 8h² - 6h) /h = (factoring h out of the numerator and simplifying) [h (16a + 8h - 6)] /h = 16a + 8h - 6 I hope it helps
SIN isa function, and the independent value of this function is a angle. A angle can be turned to a number.example: For 30°, there is sin 30°=0.5but if we know a angle can be turned to be 0.5 by sin. how much is this angle？of course it can be 30°. 390° can also(because it is 360+30°). and even 150°.so there could be another function to turn a number to angle.this function is sin^-1 can be-330°,30°,390°, 750°....... or ........-210°, 150° ,510°,870°.......
Given f (x) = 2 x² – x+ 2, find and simplify f (x + h) -f (x) over h?
ok so you want to simplify (f(x+h)-f(x))/h the first step is to plug x+h and x into f(x) so you get ((2(x+h)^2-(x+h)+2)-(2x^2-x+2))/h which is ((2x^2+4xh+2h^2-x-h+2)-(2x^2-x+2))/h yuck dont be like i used to and forget to distribute the negative sign ^_^ (2x^2+4xh+2h^2-x-h+2-2x^2+x-2)/h 2x^2 cancels, x cancels, 2 cancels, so you are left with (4xh+2h^2-h)/h if you divide out the h, 4x+2h-1 i assume you are taking the limit as h->0, which would just be 4x-1 :D yay
Is anyone good with calculus. this problem is making my brain delirious?
First you must find the derivative. f'(x)=3x^2+10x-8 Now set the function equal to zero to find the critical numbers. 3x^2+10x-8=0 (x+4)(3x-2) x=-4, 2/3 Now you have to test each interval to see if it's a minimum or a maximum. Pick a number below -4, anynumber will work but I'm going to choose -5 Plug this into your derivative. 3(-5)^2+10(-5)-8 75-50-8=17 The derivative is increasing on the interval (-infinity, to -4) Let's test our second interval (-4,2/3) I'm going to pick zero for this, because it's the easiest. 3(0)^2+10(0)-8 -8. The function is decreasing from (-4,2/3) So since you went from increasing to decreasing, you have found a relative maximum. Now test a number bigger than 2/3. Let's pick one. 3(1)^2+10(1)-8 13-8=5 The function is increasing from (2/3,infinity) Since you went from decreasing to increasing you have found a minimum at 2/3. For points of inflection, you repeat the process but with the second derivative. f'(x)=3x^2+10x-8 f''(x)=6x+10 Find the critical numbers. 6x+10=0 x=-5/3 Test the intervals. A number less than -5/3, let's pick -120192019 The answer is negative. A number bigger than -5/3 Let's say 100 The answer is positive. So you have found the x value of the point of inflection, but becuase it is a point you need the y value, which you get by plugging it back into the function. I'll leave that to you, if you get stuck just email me. :D