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Verify Rolle's theorem for the function $y=x^{2}+2, a=-2$ and $b=2$
Solution
The function $y=x^{2}+2$ is continuous in $[-2,2]$ and differentiable in $(-2,2).$
Also $f(-2)=f(2)=6$ and hence the value of $f(x)$ at $-2$ and $2$ coincide. Rolle's theorem states that there is a point $c \in(-2,2),$ where $f^{\prime}(c)=0 .$ Since $f^{\prime}(x)=2 x,$ we get $c=0 .$ Thus at $c=0,$ we have $f^{\prime}(c)=0$ and $c=0 \in(-2,2)$
Similar Questions
Let $f, g:[-1,2] \rightarrow R$ be continuous functions which are twice differentiable on the interval $(-1,2)$. Let the values of $f$ and $g$ at the points $-1.0$ and $2$ be as given in the following table:
$x=-1$ | $x=0$ | $x=2$ | |
$f(x)$ | $3$ | $6$ | $0$ |
$g(x)$ | $0$ | $1$ | $-1$ |
In each of the intervals $(-1,0)$ and $(0,2)$ the function $(f-3 g)^{\prime \prime}$ never vanishes. Then the correct statement(s) is(are)
$(A)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly three solutions in $(-1,0) \cup(0,2)$
$(B)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(-1,0)$
$(C)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(0,2)$
$(D)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly two solutions in $(-1,0)$ and exactly two solutions in $(0,2)$