Differential Calculus: point where normal is parallel to...

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Hockeyman

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1.) The point on the curve y= SqRt 2x+1 at which the normal is parallel to the line y=-3x+6

I found the derivative which is y'= (2x + 1)^(-1/2), then i set it equal to -3 which is the slope of the other line, and i tried to solve for x. the answer choices are A) 4,3 B) 0,1 C) 1, SqRt 3 D) 4,-3 E) 2, SqRt 5 Now when i try and solve for x i don't get any of those x vaules

2.) The minimum value of the slope of the curve y= x^5 + x^3 - 2x is, i found the derivative which is y'= 5x^4 + 3x^2 -2, then i found the critical points which are x=-.632 and .632, then i'm not sure where to go next.
 
Re: Differential Calculus

1. Sorry, didn't read your answer carefully enough. You took the derivative correctly. Also, you're finding the normal parallel to y = -3x + 6. So you're not setting y' = -3 ... but y' = ___ ?

2. Yes. One represents a local maximum and another the local minimum (assuming the critical points are correct). How would you determine which is which? You can always use your second derivative test or test the values of f'(x) around the two critical points. Take your pick :wink:
 
Re: Differential Calculus

Alright so for the first one i set y'= 1/3, adn i found x= 4, then i found y=3 so the answer would be (4,3) letter A?

Then for the second one i found that x=.632 is a local min using the first derivative test, now it is asking for a value so i plugged .632 into the original eqaution for x. I found that y= -.9107, now my answer choices are the following A) 0 B) 2 C) 6 D) -2 E) none of these, so i guess the answer would be none of these letter E?
 
Re: Differential Calculus

The minimum value of the slope. Look at f''(x)
 
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