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math help
Started by Atl21
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First off, I haven't taken calc yet. However, I have been practicing these sorts of problems. So if I'm wrong, someone please correct me.
F(x)= x^3-2x g'(3)=?
F'(x)= 3x^2 +2
Since g is the inverse of f, to find g(3):
3=x^3+2x .... x=1
g'(3) = 1
f(g(3)) = 5 {f'(x)= 3(1)^2 +2 = 5}
g'(3)= 1/ f'(g(3)) = 1/5
.... So I got 1/5 for the answer.
F(x)= x^3-2x g'(3)=?
F'(x)= 3x^2 +2
Since g is the inverse of f, to find g(3):
3=x^3+2x .... x=1
g'(3) = 1
f(g(3)) = 5 {f'(x)= 3(1)^2 +2 = 5}
g'(3)= 1/ f'(g(3)) = 1/5
.... So I got 1/5 for the answer.
Yes, your approach is correct.
Atl, if it helps heres another question that I was working on today:
Let f(x) = x^3 - x + 2. If h is the inverse of f, then h'(2) could be...
First of all f'(x) = 3x^2 -1
f(h'(x)) = 1/(3x^2 -1)
Solve for h'(2).
2=x^3 - x + 2
0=x(x^2-x)
0=x(x-1)(x+1) ; x=0,-1,1
f(h'(x)) = 1/(3x^2 -1); plug in the three values for x gives you 1/2 which is one of the answer choices. For x=0 there is no answer choice so your answer would be 1/2.
Hopefully, another example helps you better understand the question/solution.
f(x)= x^5+3x^3+7x+2. Given g is the inverse of f. Find g'(13).
f(x) = x^5+3x^3+7x+2.............eq 1
step 1:
first differentiate f(x),
so f'(x) = 5x^4+9x^2+7
step 2:
find the derivative for inverse function of f(which is nothing but f(g'(x)) which is 1/f'(x)
f(g'(x))= 1/(5x^4+9x^2+7)...........eq 2
step 3
Now to find x:solve for g'(13):
use eq 1
f(x)=x^5+3x^3+7x+2
13=x^5+3x^3+7x+2
.....x=1
step 4:
substitute x=1 in .......eq 2
f(g'(x)) = 1/5(1)^4+9(1)^2+7 = 1/21.
f(x) = x^5+3x^3+7x+2.............eq 1
step 1:
first differentiate f(x),
so f'(x) = 5x^4+9x^2+7
step 2:
find the derivative for inverse function of f(which is nothing but f(g'(x)) which is 1/f'(x)
f(g'(x))= 1/(5x^4+9x^2+7)...........eq 2
step 3
Now to find x:solve for g'(13):
use eq 1
f(x)=x^5+3x^3+7x+2
13=x^5+3x^3+7x+2
.....x=1
step 4:
substitute x=1 in .......eq 2
f(g'(x)) = 1/5(1)^4+9(1)^2+7 = 1/21.
How are you guys figuring out what x equals for the first and third problems? Those just seem incredibly complicated to figure out without a TI-83. Granted I haven't taken algebra in five years so I'm not that good anymore.
