Use the information to evaluate and compare (triangle)x and dy.
y=2-x^4, x=2, and (triangle)x=dx=0.01
Surely by now you know that ∆ is pronounced "delta".
At x=2,
∆y = f(x+∆x)-f(x) = (2-2.01^4)-(0) = 0.3224
dy = -4x^3 dx = -4(8)(-.01) = 0.32
I know that; don't know why I didn't use it. Oops.
How do you get 0 from f(x)?
oops. I read x^2-4, not 2-x^4.
So, that would be -14, not 0.
Sorry...
Dang dyslexia (not really, just careless)
To evaluate and compare the changes in y, denoted as (triangle)y, and x, denoted as (triangle)x or dx, we need to calculate the values of y for two different values of x, which are x and x+(triangle)x or x+dx.
Given:
- y = 2 - x^4
- x = 2
- (triangle)x = dx = 0.01
Step 1: Calculate y for x = 2
Substitute the value of x into the equation:
y = 2 - (2^4)
y = 2 - 16
y = -14
Step 2: Calculate y for x = 2 + dx
Substitute the value of x + dx into the equation:
y = 2 - ((2 + 0.01)^4)
y = 2 - (2.01^4)
y ≈ 2 - 16.08160401
y ≈ -14.0816
Step 3: Calculate (triangle)y
(triangle)y = y(x + dx) - y(x)
(triangle)y = (-14.0816) - (-14)
(triangle)y ≈ -14.0816 + 14
(triangle)y ≈ -0.0816
Therefore, the change in y, or (triangle)y, when x changes by dx = 0.01 is approximately -0.0816.