Thread: help with a few problems (don't need to answer all)

i have tried these many times and still can't get them.

1) f(x)= ((5x^3)-(5x^2)+9x-9) / (x-1) when x < 1
f(x)= (6x^2)-1x+a when x>=1

What value is a to make the function continuous at 1?

I tried making them equal to each other and putting x=1 but that didn't work.

2) (e^2x) - (7e^x) + 6 = 0

It says there are 2 solutions but I keep getting -33.1255 and that's not even one of them.

3) y=loga(x) with the subscript being a. It passes through (28,1).

What is the value of a?

I have no idea what to do on this one.

4) log2(x) + log2(x+3) = k

Solve for x in terms of k.

Again, I have no idea what to do. I can solve it if k is an integer but thats about it.

Any help is greatly appreciated.

1. Your approach would normally work, except that the function has a discontinuity at x = 1. But what is the limit of the function? Factor the numerator to yield . Then cancel the factors of and you are left with . At x = 1, this function is equal to 14, so the limit is 14, and that is the function value you want for the other part of your function.

2. Substitute , yielding . Then solve for u by factoring and identifying the roots: , so u = 1 or 6. Hence or . From you have the solution x = 0, and from you have the solution x = ln 6.

3. This is an easy one. The equation can be rewritten as . With (28, 1) on the graph, we have , or .

4. You can combine the two logarithms to yield , which can be written , or . Now apply the quadratic formula: . Recognizing that x must be a positive number in order to evaluate the logarithm, we have .

ooooooo. Those explanations were great. Thanks a lot.