Vrftjkry

Alright, so here's the deal. I need to find the interval of this derivative function:

f(x)= −5x2+12x−7

So far, I've gotten that the derivative is this:

f-prime(x)= -5x + 12

I tried to make the equation equal to zero. What happened was I got 2.4. Theoretically, anything below 2.4 should be increasing. Everything above 2.4 should be decreasing, I've even tested this.

But the site that I submit it to wants it in interval notation. That's great, but it won't read my interval notation. Am I wrong or is my notation wrong? If I am, what do I do with the derivative?

if we have $f(x)=-5x^2+12x-7$ so we get $f'(x)=-10x+12$ and now must be $-10x+12geq 0$ or $-10x+12le 0$

So your first clue should be that it is a function of degree $2$ so it will have only one maximum or minima, in this case, maxima since it is a quadratic opening downward$rightarrow a lt 0$. I am including the reason for this as well. Let us say we have a general quadratic $$ax^2 + bx + c$$ Now taking $x^2$ common we get $$x^2 Bigl(a + frac{b}{x} + frac{c}{x^2}Bigl)$$ Now to determine whether it will open upward or downward we simply see the result of $xrightarrow infty$ and so we get, $$infty^2 Bigl(a + frac{b}{infty} + frac{c}{infty^2}bigl)$$
So we get, $$infty^2 Bigl(abigl)=0$$
which obviusly says that if $alt0$ then at $$xrightarrow infty, f(x) rightarrow -infty$$ and for $agt0$ then at $$xrightarrow infty, f(x) rightarrow infty$$ This tells us that this function will be a downward opening parabola as seen in the figure below,
https://www.desmos.com/calculator/w0krov2wrl

Now to find the value of the maxima we simply set $$frac{df}{dx} = 0$$
That is, $$frac{dBigl(-5x^2 + 12x -7bigl)}{dx} = 0$$ Which gives us, $$-10x + 12 = 0$$ Or, $$x = frac{6}{5}$$ Now obviously since this is the maxima it means that after this point the function will be strictly decreasing so the interval for which it is strictly increasing is $$forall ; x lt frac6 5$$