If f(x)=2x+1 and g(x)=3x-2 then f(g(3))=?
g(3)=3(3)-2 =9-2=7 f(g(3))=f(7) =2(7)+1=14+1=15 Compare f(x) with f(g(3)) and notice that this is the same but with x=g(3), so just work that out first and use it as the value of x in f(x).
If g(x) = 2x- x +5 , then g (2) =?
hm, yeah, this might take the time to jot down out totally, yet to deliver you on the song a). Set f(x) = sqrt(x^2 + one million), g(x)= x^2 so f'(x) = x/sqrt(x^2+one million), and x^3/sqrt(x^2+one million) = f' g. Then f g' = 2x sqrt(x^2+one million) it rather is spinoff of (2/3) (x^2+one million)^3/2. observe formula b). artwork out du/dx. that's going to cut back the integrand
For the function g(x)=x^3-2x^2-1. at which of the following tangent points is the instantaneous raeof change equal to -1?
g(x) = x³ - 2x² - 1 ← this is a curve, i.e. a function g’(x) = 3x² - 4x ← this is its derivative …but the derivative is too the slope of the tangent line to the curve at x You want to obtain tangent points is the instantaneous rate of change equal to - 1. You must solve for x: g’(x) = 0 3x² - 4x = - 1 x² - (4/3).x = - 1/3 x² - (4/3).x + (2/3)² = - (1/3) + (2/3)² x² - (4/3).x + (2/3)² = 1/9 [x - (2/3)]² = (± 1/3)² x - (2/3) = ± (1/3) x = (2/3) ± (1/3) x = (2 ± 1)/3 x₁ = (2 + 1)/3 → x₁ = 1 g(x) = x³ - 2x² - 1 → when: x = 1 g(1) = 1 - 2 - 1 g(1) = 2 → Point (1 ; 2) x₂ = (2 - 1)/3 → x₂ = 1/3 g(x) = x³ - 2x² - 1 → when: x = 1/3 g(1) = (1/27) - (2/9) - 1 g(1) = (1/27) - (6/27) - (27/27) g(1) = (1 - 6 - 27)/27 g(1) = - 32/27 → Point (1/3 ; - 32/27)
If G(x) =2x - 3, would G^-1(7) = 3/2+7/2?
So the best place to start when trying to find the inverse function is take the original equation and replace [math]x[/math] with [math]G^{-1}(x) [/math]and [math]G(x)[/math] with [math]x.[/math] Sometimes it doesn’t work because the resulting expression is too difficult to handle or some other more nuanced reason. But it’s a good place to start and it works just fine here. Once we do the replacement we get [math]x=2*G^{-1}(x)-3[/math] which implies [math]G^{-1}(x)=\frac{x+3}{2}[/math]Let’s plug in 7, [math]G^{-1}(y)=\frac{7+3}{2}=5[/math] which is the answer we were looking for.
If f(x)= 2/x+3 and g(x)=1/x, then (g of f)(x) is equal to?
Work will have fractions in fractions. To hard to type but easy to skype. ANSWER is # 3 If you have questions on this or any other math problem feel free to let me know how to contact you.......Read Below....... I tutor free online since moving to France from Florida in June 2011. Hope this is most helpful and you will continue to allow me to tutor you. If you dont fully understand or have a question PLEASE let me know how to contact you so you can then contact me directly with any math questions. If you need me for future tutoring I am not allowed to give out my info but what you tell me on ***how to contact you*** is up to you Been a pleasure to serve you Please call again Robert Jones.............f "Teacher/Tutor of Fine Students" A PRESENT FOR YOU Ask someone to write a number, say a five-digit number. (Can be a 2-digit.....3-digit.....4-digit....ect..... Suppose the number written is 57836 Now, without showing the asker, you write a number on a sheet of paper and keep it folded. You have to write the number by subtracting 2 from the above number and adding 2 in front which will be.... 257834 Next, ask the person to write another five-digit number below his original number. Suppose he writes 37589. So you now have... 57836 37589 Now you write a five-digit number below it in such a way that each digit is 9 minus digit above. You now have... 57836 37589 62410 Ask the person to add one more 5-digit number. If he adds 54732, you add below it 45267. Note that you decided the number by subtracting each of his digit from 9. Thus, you now have..... 57836 37589 62410 54732 45267 Next, ask him to add all the number and he gets 257834 Show him the number which you had written as an answer to this addition earlier in the folded paper and surprise him.
What are the x and y intercepts for the function g(x)= x^3+1/ x^2+2x?
For any function the x intercepts occur when g(x) = 0, and the y-intercepts occur when we ask g(0). For example, the function f(x) = (3x^2-9)/(2x-1) has the following x- and y-intercepts: For x-intercepts f(x) = 0: f(x) = (3x^2-9)/(2x-1) = 0 ------> 3x^2 - 9 = 0. Notice the denominator vanished, why? Well we have a fraction equals a zero, which means we can multiply both sides by the denominator and boom, 0*anything = 0. Now we continue as follows Solve for x: 3x^2 - 9 = 0 3x^2 = 9 x^2 = 9/3 x^2 = 3 ---> x = +sqrt(3), x = -sqrt(3) Those are the SOLUTIONS when f(x) = 0 not the INTERCEPTS. The intercepts themselves are written as (-sqrt(3), 0), (sqrt(3), 0). Now you need to determine the y-intercepts. Find f(0). In other words, find the value of y-when x = 0. f(0) = (3*0^2 - 9)/(2*0 - 1) = (0 - 9)/(0 - 1) = -9/-1 = 9 That is, when we cross the y-axis our point is (0,9). Now, do the same sort of thing to find the values for your function. Or you could graph it and locate the points on your own.
Given f(x)=2x^2-3 and g (x)= x+2 find........?
f(x) = 2x^2 - 3 g(x) = x + 2 g(x) → g[f(x)] g(x) = x + 2 g[f(x)] = (2x^2 - 3) + 2 g[f(x)] = 2x^2 - 3 + 2 g[f(x)] = 2x^2 - 1
If g(x)=3+x+e^x, find g^-1(4)?
call g(x) = y. Then y = 3 + x + e^x. So, you want the value of the inverse function at x=4. So switch x and y to get the inverse function: x = 3 + y + e^y. So, y + e^y - 1 = 0. Then, you want the value of y that makes this true. Clearly, when y = 0, 0 + 1 - 1 = 0. I'm sure there is another way to get the answer...
Given f(x)=7x^2-3 and g(x)=1-2x find f(g(x)).?
f(x) = 7x^2-3 g(x) = 1-2x f( g(x) ) = f ((1-2x)) = 7(1-2x)^2 - 3 = 7(1 -4x+4x^2) - 3 = 7 -28x+28x^2 - 3 = 4 -28x+28x^2
Given: f(x) = x2 - 1 and g(x) = 2x + 3. Find ( G 0 F)(2)?
f(x) = x² - 1 f(2) = 2² - 1 f(2) = 4 - 1 f(2) = 3 g(x) = 2x + 3 g(f(2)) = 2([f(2)] + 3 g(f(2)) = 2(3) + 3 g(f(2)) = 6 + 3 g(f(2)) = 9 ¯¯¯¯¯¯¯¯¯