For which a and b is f(x) = ax^2 + b if x < 1; 1/x if x ≥ 1 differentiable at x = 1?

Question 124

HardOriginal question

For which $a$ and $b$ is $f(x) = \begin{cases} ax^{2} + b, & x < 1 \\ \frac{1}{x}, & x \ge 1 \end{cases}$ differentiable at $x = 1$ ?

a = \frac{1}{2},\ b = \frac{1}{2}

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a = -1,\ b = 2

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a = -\frac{1}{2},\ b = \frac{1}{2}

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a = -\frac{1}{2},\ b = \frac{3}{2}

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Show the answer and explanation

CorrectNot quiteAnswer D

Correct answer: D

Values must match at $x = 1$ : $a + b = 1$ . Slopes must match: $2a = -\frac{1}{x^{2}} = -1$ at $x = 1$ .
So $a = -\frac{1}{2}$ and $b = 1 - a = \frac{3}{2}$ .

Worth noting

$a = \frac{1}{2},\ b = \frac{1}{2}$ takes the slope of $\frac{1}{x}$ at 1 as $+1$ ; it is $-\frac{1}{x^{2}} = -1$ .

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