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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?

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  • $\begingroup$ How do you know if a function is increasing? Just by look when the derivative is positive. Now if you want your derivative to be positive, then you have to look the derivative of the derivative because it is also a function $\endgroup$ – Luis Felipe Aug 20 at 17:08
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if we have $f(x)=-5x^2+12x-7$ so we get $f'(x)=-10x+12$ and now must be $-10x+12\geq 0$ or $-10x+12\le 0$

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  • $\begingroup$ I'm having trouble using either of those equations to find anything other than the same number, which to satisfy both must be equal to x. $\endgroup$ – user3695903 Oct 22 '14 at 21:29
  • $\begingroup$ we have for example $-10x+12\geq 0$ or $12\geq 10x$ this is equivalent to $\frac{12}{10}\geq x$ or simplified to $x\le \frac{6}{5}$ $\endgroup$ – Dr. Sonnhard Graubner Oct 22 '14 at 21:34
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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 $x\rightarrow \infty$ and so we get, $$\infty^2 \Bigl(a + \frac{b}{\infty} + \frac{c}{\infty^2}\bigl)$$ So we get, $$\infty^2 \Bigl(a\bigl)=0$$ which obviusly says that if $a\lt0$ then at $$x\rightarrow \infty, f(x) \rightarrow -\infty$$ and for $a\gt0$ then at $$x\rightarrow \infty, f(x) \rightarrow \infty$$ 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, $$\frac{d\Bigl(-5x^2 + 12x -7\bigl)}{dx} = 0$$ 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 $$\forall \; x \lt \frac6 5$$

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