let f(x)=square root x. if the rate of change of at x=c is twice its rate of change at x=1, then c=?
what???? i don't get get it.
to begin with, i took the derivative if that helps which is 1/(2*sqrtx) .
f'(x) = 1/(2*sqrtx) is correct
f'(x=1) = 1/2
The problem is to find the value of a constant x=c such that the derivative is twice the value at x=1, or 1.
f'(c) = 1 = 1/(2*sqrt c)
2 sqrt c = 1
sqrt c = 1/2
c = 1/4.
1) f'(x)=1/2 times the square root of x
2) Plug in 1 as we are finding f'(1)
3) Multiply [1/2 times the square root of x] times 2 which equals 1
3) Now equate this to [1/2 times the square root of x]
4) You should get 1/4
Oh, finding constants can be a constant struggle! But fear not, I'm here to help (or at least attempt to help) with some humor.
So we need to find the value of c where the rate of change of f(x) = √x is twice the rate of change at x = 1.
Well, it seems like this problem is trying to make us go around in circles, quite literally! But let's break it down step by step and see if we can square away the answer (pun intended).
First, we find the derivative of f(x) using some math magic. And, voila, we get f'(x) = 1/(2√x). Good job on that!
Now, let's plug in x = 1 into our derivative to get f'(1) = 1/2. Easy peasy lemon squeezy!
But wait, we want the rate of change at x = c to be twice this value. Well, talk about overachievers! So we set up the equation f'(c) = 2 * f'(1).
Substituting the values we have, 1/(2√c) = 2 * 1/2. Phew, still following?
Now, let's tackle that equation by multiplying both sides by 2 to get rid of those pesky fractions: 1/√c = 1.
To isolate that squiggly square root, we square both sides of the equation: (1/√c)^2 = 1^2. Oh, yeah, we're getting squared up now!
That simplifies to 1/c = 1. And if we cross-multiply, we find that c = 1/1, which is just good old 1!
Wait, what happened to c = 1/4? Well, it seems like the answer is c = 1, my friend, unless I'm just clowning around!
But hey, always double-check your work and make sure to triple your laughter while solving math problems! Keep up the great work, math whiz!
To solve this problem, we need to find the value of x=c such that the rate of change of f(x) at x=c is twice its rate of change at x=1.
First, let's find the derivative of f(x) = sqrt(x) using the power rule. The derivative of sqrt(x) is 1/(2*sqrt(x)). So, f'(x) = 1/(2*sqrt(x)).
To find the rate of change at x=1, we plug x=1 into the derivative equation:
f'(x=1) = 1/(2*sqrt(1)) = 1/2.
Now, we want to find x=c such that the rate of change at x=c is twice the rate of change at x=1. We set up the equation as follows:
f'(c) = 1 = 2 * (1/2).
Simplifying, we have:
1/(2*sqrt(c)) = 1.
To isolate c, we can multiply both sides of the equation by 2*sqrt(c):
2*sqrt(c) * (1/(2*sqrt(c))) = 1 * (2*sqrt(c)).
The left side simplifies to:
1 = sqrt(c).
To remove the square root, we square both sides of the equation:
1^2 = (sqrt(c))^2.
This further simplifies to:
1 = c.
Therefore, the value of x=c that satisfies the condition is c=1/4.
easy 1billion compounded to the 55th decimal x times the quantum of infinity
so X?