The number of points, at which the function f(x)=max6x,2+3x^2+|x-1| |x^2-(1)/(4)|,x (- , )…

Question

The number of points, at which the function $f(x)=max{6x,2+3{x}^{2}}+|x-1|\cos |{x}^{2}-\frac{1}{4}|,x\in (-\pi ,\pi )$, is not differentiable, is

$____$ .

Answer: 3

Accepted Answer

For the part $max{6x,2+3{x}^{2}}$, the function can change its ``active'' expression only at those points where the two inside expressions become equal.

$2+3{x}^{2}=6x\Rightarrow 3{x}^{2}-6x+2=0$

$x=\frac{6\pm \sqrt{12}}{6}=\frac{3\pm \sqrt{2}}{3}$

At these two values of $x$, the max function switches between $6x$ and $2+3{x}^{2}$. Also, the derivatives of the two expressions are different there (since $\frac{d}{dx}(6x)=6$ and $\frac{d}{dx}(2+3{x}^{2})=6x$), so $max{6x,2+3{x}^{2}}$ is not differentiable at both points.

Now look at the second part $|x-1|\cos |{x}^{2}-\frac{1}{4}|$. The factor $\cos |{x}^{2}-\frac{1}{4}|$ is continuous for all $x$, and at $x=1$ it becomes $\cos |1-\frac{1}{4}|=\cos (\frac{3}{4})
e 0$.

Because $\cos |{x}^{2}-\frac{1}{4}|$ is nonzero at $x=1$, multiplying it by $|x-1|$ does not remove the ``corner'' of $|x-1|$. So $|x-1|\cos |{x}^{2}-\frac{1}{4}|$ is not differentiable at $x=1$.

Total number of points where $f(x)$ is nonDifferentiable is 3

Solutions · 1