What is the maximum vertical distance between the line $y = x + 42$ and the parabola $y = x^2$ for $−6 \le x \le 7$?
I have the answer at the back of my book but I am not sure how they got that. The book mentions somthing about derivatives.
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Hint: $$x+42-x^2=-\left(x-\frac12\right)^2+\frac{169}4$$ has a maximum at $(\frac12,\frac{169}4)$ in the interval $[-6,7]$.
Ángel Mario Gallegos
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You're missing a minus sign in front of the squared bracket – David Quinn Nov 19 '16 at 22:31
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@ÁngelMarioGallegos thank you very much! – Hidaw Nov 19 '16 at 22:38
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Let $f(x)=x+42$ and $g(x)=x^2$. Noting that, on $[-6,7]$, we have $g(x)\leq f(x)$, what you're asking for is the maximum value of the non-negative function $h(x)=f(x)-g(x)$.
Does that give you a push in the right direction?
G Tony Jacobs
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I have the solution at the back of my book but they solved it with derivatives. – Hidaw Nov 19 '16 at 22:23
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Using derivatives is a usual way for finding the maximum value of a differentiable function. What is the value of the derivative when you're at the maximum value? – G Tony Jacobs Nov 19 '16 at 22:24
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If you want to solve it with derivatives, you first have to think about how $x^2$ behaves. You see, the slope (which is the derivative of the function) of $f(x)=x+42$ is $1$. The slope of $g(x)=x^2$ is not constant, but you can calculate it for a given point by calculating its derivative $g'(x)=2x$ and plugging the $x$-coordinate of the point in.
The derivative of $g(x)$, $g'(x)=2x$ is inferior to that of $f(x)$, $f'(x)=1$ until a certain x-coordinate, and this coordinate gives us the maximal vertical distance between the two functions.
Establish the equality:$$f'(x)=g'(x)\Rightarrow 1=2x\Rightarrow x=\frac{1}{2}$$
Plug the x value in both of the functions to find the maximal distance. $$f\left(\frac{1}{2}\right)-g\left(\frac{1}{2}\right)=\frac{170}{4}-\frac{1}{4}=\frac{169}{4}$$
Glycerius
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