I have to use the Mean Value Theorem to find f'(c)
1. x^3+x on [1,2] the answer says c=1.5275...
2. x^4+2 on[-1,2] the answer says c=1.0772...
The entire directions state: Verify that the given function f satisfies the hypothesis of the MVT on the given interval [a,b]. Then find all numbers cbetween a and b for which: (f(b)-f(a)/b-a)=f'(c)
For 1, I found the deriv. which is 3x^2+1
Then applied 1 and 2 to the original equation and got 4 and 13 respectively.
plugged them into the MVT equation:13-4/2-1=9.
Then I plugged the 9 with the derivitive 3x^2+1 for it to become 3x^2+1-9=3x^2-8.
Made x stand by itself and got x = +or-(sqrt 8/3). and it doesn't add up to the answer.
For 2, Long story short, I did the exact same steps as in #1 and the result was the same.
1/x+1 on [0,2]. the answer says .73...
The deriv. is -1/(x+1)^2 with the quotient rule. f(0)=1 and f(2)=1/3 then MVT , the answer came as -2/3/2 or -2/6=-1/3.
I plugged that with the derivitive: -1/(x+1)^2+1/3.
I made them like to get rid of the denominators to get -3+x^2+1.
x^2=2
x=+or-(sqrt 2)
And I know I did not get this right again.
Oh yeah, disregard 2. I got it.
I got all of that. The problem is getting the c to be alone since it is (x+1)^2 I am working with. I have attempted quadratic but it did not work.
I have -1/(c+1)^2=-1/3
mult. the (c+1)^2 to be -1=-1/3(c+1)^2
I broke the binomial for the right side to be (-1/3)c^2-(2/3)c+1.
put 1 on other side: -2=(-1/3)c^2-(2/3)c
-2=(c-1)(-1/3)c
I know what your trying to say though, it is the (c+1)^2 that is in the way.