Please help me with this WS
1) Find the value of lim as x-->-(square root of 3) of (-x^4 + 5x^2 -6)/(x+(square root of 3)
2) Find the distance between the discontinuities of g(x)= ((square root of x-4))/(x^2 -5x -1)
3)Find the value of x at which the curve y=(x^2 - 4)/(x^5 + 2x^4 + 6x + 12) has a Vertical Asymptote
4)lim as x-->-2 of (1)/(x+2) + (4)/(x^2 -4)
5)The volume of a cube is increasing at the rate of 20 cubic cm per second. How fast, in square cm per second, is the surface area of the cube increasing at the instant when the edge of the cube is 10 cm long?
6)The normal line to the curve y=(square root of 16-x) at the point of (0,4) has a slope of what?
-(x^4 - 5 x^2 + 6) / (x +3^.5)
-(x^2-3)(x^2-2) /(x+3^.5) OH LOOK!
-(x-3^.5)(x+3^.5)(x^2-2)/(x+3^.5)
= -(x-3^.5)(x^2-2)
when x is -3^.5
= -(-2*3^.5)(1)
= 2 sqrt 3
Now there is a limit and I am at it without you doing some.
Sure, I can help you with these questions. Let's go through each one and explain how to solve them.
1) To find the value of the limit as x approaches -(square root of 3) of (-x^4 + 5x^2 -6)/(x+(square root of 3)), you can perform direct substitution. Simply substitute the value -(square root of 3) into the expression and evaluate it.
2) To find the distance between the discontinuities of g(x) = ((square root of x-4))/(x^2 -5x -1), you need to identify the values of x for which the function is undefined. These are the points of discontinuity. To do this, set the denominator equal to zero and solve for x. The difference between the two values of x will give you the distance between the discontinuities.
3) To find the value of x at which the curve y = (x^2 - 4)/(x^5 + 2x^4 + 6x + 12) has a vertical asymptote, you need to check where the denominator of the function becomes zero. Set the denominator equal to zero and solve for x. The value(s) that make the denominator zero will determine the vertical asymptote.
4) To find the limit as x approaches -2 of (1)/(x+2) + (4)/(x^2 -4), you can simplify the expression first. Rewrite (4)/(x^2 -4) as (2)/(x+2) - (2)/(x-2) and then combine it with (1)/(x+2). Now you have a new expression, which can be evaluated by direct substitution.
5) To find how fast the surface area of a cube is increasing when the volume is increasing at 20 cubic cm per second and the edge length is 10 cm, you need to use the derivative. The surface area of a cube is given by 6 times the square of the edge length. Differentiate the surface area formula with respect to time and substitute the given values into the resulting expression.
6) To find the slope of the normal line to the curve y = (square root of 16-x) at the point (0,4), you need to find the derivative of the curve and evaluate it at x = 0. The derivative will give you the slope of the tangent line, but the normal line will have a slope that is the negative reciprocal of the tangent line slope. So, take the negative reciprocal of the slope obtained from the derivative evaluation at the given point.