Recent content by sp09ta

Hehehe... I had my value for b-a incorrect, which threw off my whole solution.. :D Got the right answer now, thanks for the feedback ;)

1. Verify that the function satisfies the mean value theorem, then find all numbers c that satisfy the the conclusion of the mvt f(x)=e^(-2x) on the interval [0,3] 2. f'(c)=[f(b)-f(a)]/[b-a] 3. 1. f(x) is a composition of continuous functions, so f(x) is continuous on[0,3]. 2. f(x) is a...

Ok, then f(1)=a+b+4=-1 and f'(1)=3a+2b+6=-3 so, a+b+5=0 and 3a+2b+9=0 then I isolate b in the first equation and sub it into the second to get: 3a+2(-a-5)+9=0 a=1 Then use a to find the value of b So my values are: a=1, b=-6, c=2, d=1

1. Suppose the curve f(x)=(x^4)+a(x^3)+b(x^2)+cx+d has tangent line when x=0 with equation y=2x+1, and a tangent line when x=1 with equation y=2-3x. Find a,b,c,d 2. d/dx f(x)g(x)=g*f`+f*g` 3. f`(x)=4x^3+3ax^2+2bx+c x=0 y=2x+1 y=1 (0,1) is a point on f(x) x=1 y=2x-3 y=-1 (1,-1) is a point on...

K its working perfectly now that I used the round parenthesis instead of square! Thx mathie, won't make that mistake again!

Oh. Maybe it's because I was using Maple v.13. I'll retry on v.11.

1. I have 2 problems in which I'm asked to plot a function using maple(sorry i don't know latex yet): a. f(x)=[x^3-1]/[sqrt(x)-1] - find the lim x->1 Here i receive an empty graph for some reason... :( b. f(x)=[tan(3x)]/[tan(5x)] - find the lim x->0 When i attempt to plot this...

Oh, I guess my theory on the geometry of inverses was a little off. I write exp(1)^p because its the maple command that I was using. Thanks for your comment!

1. Plot f(x)=5+ln(-2x+3), and its inverse. 2. I know the inverse of ln(x) is exp(1)^(x)., and to find the inverse of a function, solve for x then swap the x and y. 3. y=5+ln(-2x+3) y-5=ln(-2x+3) exp(1)^(y-5)=(-2x+3) (-1/2)*[exp(1)^(y-5)-3]=x...