Let g(t)=(0.7)^t. Estimate g′(7) to within one decimal place by using a small enough interval.
g′(7)≈
(If you can show work please do!! trying to understand the whole problem if possible! Thanks!)
so pick ∆x = 0.01
That means that a good approximation for f'(7) would be
(f(7.01)-f(7))/(7.01-7)
now just crank 'er out.
Thank you!!! Got it!
To estimate g′(7), we need to find the derivative of the function g(t)=(0.7)^t and evaluate it at t=7.
First, we can find the derivative of g(t) using the exponential rule:
g'(t) = ln(0.7) * (0.7)^t
Then, we can evaluate g'(t) at t=7:
g'(7) = ln(0.7) * (0.7)^7
To estimate g′(7) to within one decimal place, we can use numerical approximation methods like the finite difference method or Taylor series approximation.
Let's use the finite difference method:
To use the finite difference method, we need to choose a small enough interval h. Let's try h=0.1.
Using the formula for the finite difference method:
g'(7) ≈ (g(7+h) - g(7))/h
Substituting the values into the formula:
g'(7) ≈ ((0.7)^(7+0.1) - (0.7)^7) / 0.1
Now we can compute this:
g'(7) ≈ ((0.7)^7.1 - (0.7)^7) / 0.1
Calculating this expression gives the estimated value of g′(7) to within one decimal place.