Science:Math Exam Resources/Courses/MATH110/December 2010/Question 07

MATH110 December 2010

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Question 07
A function $\displaystyle f$ is defined piecewise by $f(x)={\begin{cases}-x^{2}&{\hbox{if }}x<1\\ax^{3}+bx-2&{\hbox{if }}x\geq 1\end{cases}}$ Find values of a and b so that $\displaystyle f$ is continuous and differentiable everywhere.

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Hint
A function is continuous at a point x = a if its limits from both sides of a exist and the value of these limits agrees with $f(a)$. Furthermore, the function is differentiable at a if the derivatives from both sides of a exist and agree at a.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. It is clear that this function is continuous and differentiable everywhere, except maybe at x = 1. Let's start with checking continuity. For f(x) to be continuous at x = 1 the left and right limits of the function must be the same. Hence we set ${\begin{aligned}\lim _{x\to 1^{-}}f(x)&=\lim _{x\to 1^{+}}f(x)\\\lim _{x\to 1}-x^{2}&=\lim _{x\to 1}ax^{3}+bx-2\end{aligned}}$ Because these functions are polynomials we can simply substitute the x-value to get: ${\begin{aligned}-(1)^{2}&=a(1)^{3}+b(1)-2\\-1&=a+b-2\end{aligned}}$ This gives us one equation in terms of a and b. We can't proceed any further with this equation right now, so we will turn to the second condition that the function be differentiable. This means that the derivatives from the right and left of x = 1 must be equal. In other words: $\displaystyle (-x^{2})'=(ax^{3}+bx-2)'$ at $\displaystyle x=1$ $\displaystyle -2x=3ax^{2}+b$ at $\displaystyle x=1$ Plugging in x = 1 this yields $\displaystyle -2(1)=3a(1)^{2}+b$ $\displaystyle -2=3a+b$ which gives another equation in terms of a and b. We know both equations must hold if $f(x)$ is to be continuous and differentiable, so we can solve them as a system of equations to find a and b. Solving the first equation for b we get: $\displaystyle b=1-a$ Substituting this into the second equation we get: $\displaystyle {\begin{aligned}-2&=3a+1-a\\-3&=2a\\{\frac {-3}{2}}&=a\end{aligned}}$ Plugging this back into $b=1-a$ we get that $b={\frac {5}{2}}$. (Note that the system of equations can be solved multiple ways - using a different substitution or elimination would also work.) So $a=-{\frac {3}{2}}$ and $b={\frac {5}{2}}$ are values that satisfy the conditions that $f(x)$ be continuous and differentiable.

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