Estimating f(x) & Computing dy/dx: Answers vs. Solutions

[buffgilville]

1) Use differentials to estimate f(8.05) given that f(8)=71 and
(d/dx)f(x)=sqrt of (x+56)

i used f(c+h) = f(c) + fprime(c) * h
i got 8.125, but the answer is 71.40

2) Compute dy/dx at the given values xy - 6y^2 - 2x = -24 , x=1, y=2

I keep getting 4/25
but the answer is 0

3) Give a value c that satisfies the conclusion of the mean value theorem on the interval 2 < x < 5 for the function f(x) = 6x^2 + 8x + 13 (note: all < means greater than or equal to )

I took the prime of f(x), and f '(c) = 12x + 8
12x + 8 = 5
and I got c= (-1/4)
but the answer is c = (7/4)

 

[Fredrik]

1. The formula is correct, and when I'm using it, I get 71.4.
2. The question doesn't make sense.
3. f'(c) is 50, not 5.

 

[cepheid]

I think 2) should read:

For the following expression: xy - 6y^2 - 2x = -24

Compute dy/dx for the following values: x = 1, y = 2

Differentiate implicitly wrt x:

[tex] xy - 6y^2 - 2x = -24 [/tex]

[tex] [x\frac{dy}{dx} + y] - 12y\frac{dy}{dx} - 2 = 0 [/tex]

[tex] \frac{dy}{dx}(x - 12y) = 2 - y [/tex]

[tex] \frac{dy}{dx} = \frac{2 - y}{x - 12y} = \frac{2 - 2}{1 - 24} =0 [/tex]

 

[Fredrik]

OK, that makes sense. I messed up and thought that (1,2) isn't even on that curve, but I guess 2-24-2 really is -24.

 

[HallsofIvy]

1) Use differentials to estimate f(8.05) given that f(8)=71 and
(d/dx)f(x)=sqrt of (x+56)

i used f(c+h) = f(c) + fprime(c) * h
i got 8.125, but the answer is 71.40

I wish you had shown your work. You seem to be making some serious mistakes. In this case, you are told that f(c)= f(8)= 71 so I don't see how you could haved gotten 71+ something equal to "8.125".
Yes, f(c+ h)= f(c)+ f'(c)*h (approximately)
so f(8+ .05)= f(8)+ f'(8)*(0.05)
= 71+ sqrt(8+56)*(0.05)
= 71+ sqrt(64)*(0.05)
= 71+ 8*(0.05)= 71+ 0.4= 71.4

2) Compute dy/dx at the given values xy - 6y^2 - 2x = -24 , x=1, y=2

I keep getting 4/25
but the answer is 0

HOW do you get "4/25"? Using "implicit differentiation",
y+ xy'- 12yy'- 2= 0. Setting x= 1, y= 2, this is
2+y'- 24y'- 2 or -23y'= 0. I wonder if you weren't getting the signs mixed up?

3) Give a value c that satisfies the conclusion of the mean value theorem on the interval 2 < x < 5 for the function f(x) = 6x^2 + 8x + 13 (note: all < means greater than or equal to )

I took the prime of f(x), and f '(c) = 12x + 8
12x + 8 = 5
and I got c= (-1/4)
but the answer is c = (7/4)

You took the derivative of f(x)! Don't use "slang"!

The mean value theorem says [itex]\frac{f(b)-f(a)}{b- a}= f '(x)[/itex] for some x between a and b. In this case, a= 2, b= 5 so f(b)= 6(25)+ 8(5)+ 13= 150+ 40+ 13= 203 and f(a)= 6(4)+ 8(2)+ 13= 24+16+ 13= 53. We are looking for c so that [itex]f '(c)= 12c+ 8 = \frac{203-53}{5-2}=\frac{150}{3}= 50. 12c+ 8= 50 so 12c= 42. c= 42/12= 7/2 (?Not 7/4).
I have no idea where you got "5". If 12c+ 8= 41, then 12c= 33

 

[buffgilville]

Thanks for everyone's help!