Question

If f(x) = x3 - 2px2 - 4x + 5 and f'(2) = 0, find p.
PLEASE Answer!!​

Answer 1

Answer:

f(x) = x³-2px²-4x+5

f(2) = 0

so,

(2)³-2p(2)²-4(2)+5 = 0

8-2p(4)-8+5 = 0

-8p+5 = 0

-8p = 5

p = -5/8

Answer 2

Answer:

p = 1.

Step-by-step explanation:

f(x) = x³ - 2px² - 4x + 5

and f¹(2) = 0. Here f¹ implies  $\frac{d}{dx}(f(x))$.

Let's differentiate f(x) with respect to x so that we can find f¹(x).

To do so remember this $\frac{d}{dx} (x^n) = nx^n^-^1$.

Also $\frac{d}{dx}(CONSTANT) = 0$.

If you want to differentiate $ax^n$, you have to do this,

$\frac{d}{dx}(ax^n)=a\frac{d}{dx}(x^n)=anx^n^-^1$.

Therefore $\frac{d}{dx}(f(x)) =$ f¹(x) = $\frac{d}{dx}(x^3 - 2px^2 -4x+5)$

= $\frac{d}{dx}(x^3) - 2p\frac{d}{dx}x^2-4\frac{d}{dx}x+\frac{d}{dx}5$

= 3x² - 4px - 4.

Therefore f¹(x) = 3x² - 4px - 4.

f¹(2) = 0

i.e 3(2)² - 4p(2) - 4 = 0

12 - 4 -8p = 0

8p = 8

p = 1.

P.S : I have did a mistake previously (the answer was deleted), sorry for that.

I hope you meant f¹(2) = 0, because that is what written in the question, not f(2) = 0.