Question

need help on the attachment @Calculus1

Answer 1

separate variables and integrate both sides?

Answer 2

could I use the product rule and then use point slope form to find the y-intercept?

Answer 3

essentially, yeah. The indefinite integral will yield a +c that you'll have to solve for using the point (2,2), then you'll have a function you can evaluate at x=0.

Answer 4

could you show me your way because we I did it my way I got 4x^(2)+2+4x/sqrt(2x^(2)+1) for the slope after the using the product rule

Answer 5

???

Answer 6

I used u-substitution: let u = 2x^2 +1, so du = 4xdx
2xdx = (1/2)du, so \[f = \int\limits_{}^{}(1/2)u ^{1/2} du\]

Answer 7

what your a and b endpoints

Answer 8

Also how did you get 2xdx shouldn't it just be 4xdx?

Answer 9

No endpoints, indefinite integral. \[f = (1/3)(2x^2 +1)^{3/2} +c\]
du = 4xdx, but that is 2*(2xdx), so the original slope function has (1/2)du.

Answer 10

so how would you find the y-intercept from that would you just solve for the c variable?

Answer 11

Now substitute (2,2) for (x,f) to solve for c. Then with that complete formula, sub. in x=0 to get y-intercept.

Answer 12

did you get -7

Answer 13

I'm double-checking the derivative now.

Answer 14

c=-7, but that isn't necessarily the y-intercept.

Answer 15

then what is?

Answer 16

The final equation for f is \[f = (1/3)(2x^2 +1)^{3/2} -7\]
Evaluate at x=0, and y=f= -20/3

Answer 17

If you have a graphing utility like a TI-83/84 calculator, you can verify the points, and the slope.

Answer 18

so is the y intercept -20/3?

Answer 19

Yep. That is also its minimum value. The slope there is zero.