To find f'(3) (f prime of 3), you must find f' first. f' is the derivative of the function f(x).
Finding the derivative of f(x) = 2x⁴ requires the use of the power rule.
The power rule for derivatives is ![\frac{d}{dx} [x^{n} ] =nx^{n-1}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%20%5Bx%5E%7Bn%7D%20%5D%20%3Dnx%5E%7Bn-1%7D)
. In other words, you bring the exponent forward and multiply it by the coefficient of the term, and then you subtract 1 from the original exponent.
f'(x) = 
(2x⁴)
f'(x) = 2(4)x³
f'(x) = 8x³
Now, to find f'(3), plug 3 into your derivative.
f'(3) = 8(3)³
f'(3) = 216
<h3>Answer:</h3>
f'(3) = 216
If you have a quadratic equation in the form ax^2+bx+c
Step 1) Determine the product of AC (the coefficients in a quadratic equation)
Step 2) Determine what factors of a⋅ca⋅c sum to bb
Step 3) "ungroup" the middle term to become the sum of the factors found in step 2
Step 4) group the pairs.
Step-by-step explanation:
We have
First, 125 is a perfect cube because
and
x^3 is a perfect cube because
so we can use the difference of cubes identity
Let say we have two perfect cubes:
64 because 8×8×8=64
and 27 because 3×3×3=27 and let subtract
we know that
but using the difference of cubes identity we should get the same thing.
Remeber cube root of 64 is 4 and cube root of 27 is 3 so we have
So the difference of cubes works for real numbers. This is a good way to help remeber the identity using real numbers.
Back on to the topic,
we know that 5 is cube root of 125 and x is the cube root of x^3 so we have
You have to find 1/3 of 24, so it is 8
