please help! having issue with exponent part of problem. I come to 16x^4 but I think it should be 16x^3.i also have complications with finding the domains.
find the derivative of the function using the definition of derivative.
f(x)=4x^4
f'(x)=?
State the domain of the function.
state the domain of its derivative.
f(x+h) = 4(x+h)^4 = 4(x^4+4x^3h+6x^2h^2+4xh^3+h^4
f(x+h)-f(x) = 4(4x^3h+6x^2h^2+4xh^3+h^4)
(f(x+h)-f(x))/h = 4(4x^3+6x^2h+4xh^2+h^3)
now take the limit as h->0 and you have 16x^3
the domain of any polynomial is all real numbers
To find the derivative of a function using the definition of derivative, we use the formula:
f'(x) = lim(h->0) [f(x + h) - f(x)] / h
Let's start by finding the derivative of your function f(x) = 4x^4.
Step 1: Substitute the function into the derivative definition formula:
f'(x) = lim(h->0) [4(x + h)^4 - 4x^4] / h
Step 2: Expand and simplify the expression inside the limit:
f'(x) = lim(h->0) [4(x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4) - 4x^4] / h
= lim(h->0) [4x^4 + 16x^3h + 24x^2h^2 + 16xh^3 + 4h^4 - 4x^4] / h
= lim(h->0) [16x^3h + 24x^2h^2 + 16xh^3 + 4h^4] / h
= lim(h->0) 16x^3 + 24x^2h + 16xh^2 + 4h^3
Step 3: Cancel out the h from each term that has an h in its denominator:
f'(x) = 16x^3 + 24x^2h + 16xh^2 + 4h^3
= 16x^3 + 24x^2h + 16xh^2 + 4h^3
Step 4: Take the limit as h approaches 0:
f'(x) = 16x^3 (as h->0)
Therefore, the derivative of f(x) = 4x^4 is f'(x) = 16x^3.
Now let's discuss the domain of both the function and its derivative.
The domain of a function is the set of all values of x for which the function is defined. In the case of f(x) = 4x^4, the domain is all real numbers. There is no restriction on x since any real number can be raised to the power of 4.
The domain of the derivative, f'(x) = 16x^3, is also all real numbers. Since taking derivatives does not introduce any new restrictions, the domain of the derivative is the same as the domain of the original function.
In summary:
- The derivative of f(x) = 4x^4 is f'(x) = 16x^3.
- The domain of f(x) = 4x^4 is all real numbers.
- The domain of f'(x) = 16x^3 is also all real numbers.