# Thread: y(x) = (5-x)^4 at x = 4 can anyone help me solve this with the intantaneous/rate form

1. ## y(x) = (5-x)^4 at x = 4 can anyone help me solve this with the intantaneous/rate form

y(x) = (5-x)^4 at x = 4
how do I solve this equation with the instantaneous rate of change formula ? any help would be greatly appreciated .I know I use f'(x)=lim{f((x+h)-f(x))/h) but the polynomial is giving me trouble ....please help ..the answer is -4 from a quiz I did but I got it wrong .
Bee

2. ## Re: y(x) = (5-x)^4 at x = 4 can anyone help me solve this with the intantaneous/rate

Originally Posted by bee77
y(x) = (5-x)^4 at x = 4
how do I solve this equation with the instantaneous rate of change formula ? any help would be greatly appreciated .I know I use f'(x)=lim{f((x+h)-f(x))/h)
Do you understand that $y(x) = (5-x)^4=(x-5)^4~?$ If you do not there is no point in your trying this question.
To answer you must know that the instantaneous rate of change formula is simply: $y'(x)=4(x-5)^3$.
So $y'(4)=-4$.

3. ## Re: y(x) = (5-x)^4 at x = 4 can anyone help me solve this with the intantaneous/rate

Thank you . So everytime I see something like this I just simply derive with the power rule and substitute?If a limit was added would I need to refer to the f'(x)=lim{f((x+h)-f(x))/h) formula?

4. ## Re: y(x) = (5-x)^4 at x = 4 can anyone help me solve this with the intantaneous/rate

sorry the limit ->0 should have been added then

5. ## Re: y(x) = (5-x)^4 at x = 4 can anyone help me solve this with the intantaneous/rate

Not quite...

$\displaystyle f'(x) = \lim_{h \to 0} \dfrac{(f(x+h)-f(x))}{h}$

$\displaystyle f'(x) \neq \lim_{h \to 0} \dfrac{f((x+h)-f(x))}{h}$

Placement of parentheses is important.

6. ## Re: y(x) = (5-x)^4 at x = 4 can anyone help me solve this with the intantaneous/rate

$\displaystyle f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}$

$\displaystyle f(x) = (5-x)^4$, $f(x+h) = [5-(x+h)]^4 = (5-x-h)^4$

let $\displaystyle t = 5-x$

$\displaystyle \lim_{h \to 0} \dfrac{(t-h)^4 - t^4}{h}$

$\displaystyle \lim_{h \to 0} \dfrac{[(t-h)^2 - t^2][(t-h)^2 + t^2]}{h}$

$\displaystyle \lim_{h \to 0} \dfrac{(t^2 - 2ht + h^2 - t^2)(t^2 - 2ht + h^2 + t^2)}{h}$

$\displaystyle \lim_{h \to 0} \dfrac{h(h-2t)(2t^2-2ht+h^2)}{h}$

$\displaystyle \lim_{h \to 0} \dfrac{\cancel{h}(h-2t)(2t^2-2ht+h^2)}{\cancel{h}} = -4t^3 = -4(5-x)^3$

$\displaystyle f'(4) = -4(5-4)^3 = -4$

7. ## Re: y(x) = (5-x)^4 at x = 4 can anyone help me solve this with the intantaneous/rate

Originally Posted by bee77
y(x) = (5-x)^4 at x = 4
how do I solve this equation with the instantaneous rate of change formula ? any help would be greatly appreciated .I know I use f'(x)=lim{f((x+h)-f(x))/h) but the polynomial is giving me trouble ....please help ..the answer is -4 from a quiz I did but I got it wrong .
Bee
The way you have written this is very confusing! You say "solve this equation" but the given equation has two variables so that "solve the equation" makes no sense. And, it appears that you really want to "find the derivative of this function at x= 4. In any case, you should have shown us what you did and why you got "-4" so we could point out and correct any error.