Question
Verify Rolle's theorem for the following function on the indicated intervals $f(x) = x(x - 1)^2$ on $[0, 1]$
✓
Answer
Given $f(x) = x(x - 1)^2$
$\Rightarrow f(x) = x(x^2 - 2x + 1)$
$\therefore f(x) = (x^3- 2x^2+ x)$
We know that a polynominal function is everywhere derivable and hence continuous.
So, being a polynomial function$, f(x)$ is continuous and derivable on $[0, 1]$
Also,
$f(0) = f(1) = 0$
Thus, all the continuous of Solids theorem are satisfied.
Now, we have to show that there exists $\text{c}\in(0,1)$ such that $f\ '(c) = 0.$
We have
$f(x) = x^3 -2x^2 +^{ }x$
$\Rightarrow f\ '(x) = 3x^2 - 4x + 1$
$\therefore f\ '(x) = 0$
$\Rightarrow 3x^{2 }- 4x + 1 = 0$
$\Rightarrow 3x^{2 }- 3x - x + 1 = 0$
$\Rightarrow 3x(x - 1) - 1(x - 1) = 0$
$\Rightarrow (x - 1)(3x - 1) = 0$
$\Rightarrow\text{x}=1,\frac{1}{3}$
Thus, $\text{c}=\frac{1}{3}\in(0,1)$ such that $f\ '(c) = 0.$
Hence, Rolle's theorem is verified.
$\Rightarrow f(x) = x(x^2 - 2x + 1)$
$\therefore f(x) = (x^3- 2x^2+ x)$
We know that a polynominal function is everywhere derivable and hence continuous.
So, being a polynomial function$, f(x)$ is continuous and derivable on $[0, 1]$
Also,
$f(0) = f(1) = 0$
Thus, all the continuous of Solids theorem are satisfied.
Now, we have to show that there exists $\text{c}\in(0,1)$ such that $f\ '(c) = 0.$
We have
$f(x) = x^3 -2x^2 +^{ }x$
$\Rightarrow f\ '(x) = 3x^2 - 4x + 1$
$\therefore f\ '(x) = 0$
$\Rightarrow 3x^{2 }- 4x + 1 = 0$
$\Rightarrow 3x^{2 }- 3x - x + 1 = 0$
$\Rightarrow 3x(x - 1) - 1(x - 1) = 0$
$\Rightarrow (x - 1)(3x - 1) = 0$
$\Rightarrow\text{x}=1,\frac{1}{3}$
Thus, $\text{c}=\frac{1}{3}\in(0,1)$ such that $f\ '(c) = 0.$
Hence, Rolle's theorem is verified.
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