Solve $f(x) = ax^2 + bx + c$ to find the value of $K$

Question

$f(x)=ax^2+bx+c$, where $a=-9$, $b=12$ and $c=16$. If $$-1<f'(x)<1$$ then $h<x<k$. To $2$ decimal places, what is the value of $k$?

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Hi, this is working for solving $f(x) = ax^2 + bx + c$ to find the value of $K$

I used Quadratic Equation to solve for x and got the answer for

$x = -0.824045318$ or $2.157378652$

Hence, the value of K is -0.82

Did I used the correct formula to solve this question?

Answer 1

well, suppose $f(x) = ax^2 + bx + c $, then $f'(x) = 2ax + b $. If $-1 < f'(x) < 1$, then

$$ -1 < 2ax + b < 1 \iff -b-1 < 2ax < 1 - b \iff - \frac{b+1}{2a} < x < \frac{1-b}{2a}$$

Answer 2

Read the question carefully. You are not asked to find the roots of the quadratic equation (which you correctly found with the quadratic equation), instead you are asked to find the $k$. We have that $$f(x) = -9x^2+12x+16 $$ so $$f'(x)=-18x+12. $$ It is given for $h<x<k$ that $$-1<f'(x)<1 $$ in which we plug our expression for $f'(x)$ in \begin{align} -1<-18x+12<1\implies &\text{(subtract 12)} \\ -13<-18x<-11 \implies &\text{(divide by $-18$)}\\ \frac{-13}{-18}>x>\frac{-11}{-18}\implies&\text{(simplify)} \\ \frac{11}{18}<x<\frac{13}{18}. \end{align}

We changed the direction of the inequalities because we divided by the negative number $-18$. We can now see that from $h<x<k$ that $$k=\frac{13}{18}\approx 0.72.$$