Add the five to each side.
then square it to get rid of the square root
2x+13=(5+x)^2
2x+13=x^2+10x+25
2x=x^2+10x+12
x^2+8x+12=0
(x+6)(x+2)
x+6=0
x+2=0
answers are -2 and -6, however, when pluging -6 in it doesnt work, so it is only -2.
Given: f(x)=(2x-2)/4
find f^-1(3)?
we need to find the inverse of f(x), so
x=(2y-2)/4
2y-2=4x
y-1=2x
y=2x+1
so, then f^-1(x)=2x+1
f^-1(3)=2(3)+1
=6+1
=7
so, the answer is 7
Answer:
a) The function is constantly increasing and is never decreasing
b) There is no local maximum or local minimum.
Step-by-step explanation:
To find the intervals of increasing and decreasing, we can start by finding the answers to part b, which is to find the local maximums and minimums. We do this by taking the derivatives of the equation.
f(x) = ln(x^4 + 27)
f'(x) = 1/(x^2 + 27)
Now we take the derivative and solve for zero to find the local max and mins.
f'(x) = 1/(x^2 + 27)
0 = 1/(x^2 + 27)
Since this function can never be equal to one, we know that there are no local maximums or minimums. This also lets us know that this function will constantly be increasing.
Answer:
x = 0
x = 3
Step-by-step explanation:
