Russ Math

[Russ Math @russ_math]

For 3 on the HW, I think that "at rest" means v(t)=0. Since the object is not in motion, it is not moving and has no velocity, ergo at rest

[Russ Math @russ_math]

For final velocity problems, I looked up the formula, but how are we supposed to know t's value when it hits the ground? @STAB_MATH #noinfo

[Russ Math @russ_math]

I'm referring specifically to #3 on the homework @STAB_MATH

[Russ Math @russ_math]

How does the product rule apply if g(x) has a constant? The derivative of g(x)=3 is 0, which would nullify the equation, so that's not right

[Russ Math @russ_math]

@claireSTABmath Simplify the denom to x^(2/3) using a rule of exponents. Then transfer the x-value to the num and make the exponent -2/3, should set u up.

[Russ Math @russ_math]

For #2, I got 2.5. Use the power rule, which leads to f'(x)=2x-3. Then set 2x-3 equal to 2, which gets you 2.5.

[Russ Math @russ_math]

For a root sign in the denom you take the root # and divide the exponent inside the root by it, then flip the +/- and put it in the num

[Russ Math @russ_math]

If f'(x) is a line (pos slope), then f(x) is parabola that opens up since the y-val of f'(x) go from neg to pos

[Russ Math @russ_math]

@STAB_MATH On #5, f(x) will be a parabola. The vertex will be 1, 0.5 since f'(x)=0 there. Therefore, slope is + when x>1 and - when x<1

[Russ Math @russ_math]

For 2c, I said g(x) does reach +, since after (0,0), we can deduce that the slope drops to around -2, then cancels out by reaching 2 again?

[Russ Math @russ_math]

I need some help. For an absolute value graph f(x)=IxI, is the f'(x) graph a straight line at y=-1, then a break at 0, then straight y=1?

[Russ Math @russ_math]

For #1, I'm getting that the cubic func is f(x). For f'(x), it's parabolic since the slope f(x) gets exponentially shorter then increases

[Russ Math @russ_math]

So, the pattern is when f(x) is cubic, f'(x) is a parabola; when f(x) is parabola, f'(x) is a line w/ a slope, and so on; right?

[Russ Math @russ_math]

On 2nd 6c, I ended up with i√13 in the slope's denominator, since when you plug -2 into f'x (for which I got 2/√4x-5), you end up w/ a neg #

[Russ Math @russ_math]

On the first 6c, I have it down to f'x = (3/(2√3x-1)) but doesn't come out cleanly and I cannot deduce the slope to solve f(3) @STAB_MATH

[Russ Math @russ_math]

For #11, I'm getting 7 for both responses. The function is linear, so the slope is constant and therefore no derivative is unique.

[Russ Math @russ_math]

IDK about you, but I think the derivative is 0 at the max/min point of a parabola-shaped graph (equal +/- slope on either side)

[Russ Math @russ_math]

I think a derivative is an instantaneous slope -- that is, the slope of a function at a given point (not overall). What about you?

[Russ Math @russ_math]

I found the common denominators up top, then subtracted and multiplied by 1/denom. What about you?

[Russ Math @russ_math]

What are y'all doing for this one? I did the rule of cubes and I get that b=0 in the denom but I can't simplify anything further.