$\text{f}(\text{x})=\sqrt{25-\text{x}^2}$
Here, f(x) will exist, if
$25-\text{x}^2\geq0$
$\Rightarrow\text{x}^2\leq25$
$\Rightarrow-5\leq\text{x}\leq5$
Since for each $\text{x}\in[-3,4],$ the function f(x) attains a unique definite value.
So, f(x) is continuous on [-3, 4]
Also, $\text{f}'(\text{x})=\frac{1}{2\sqrt{25-\text{x}^2}}(-2\text{x})=\frac{-\text{x}}{\sqrt{25-\text{x}^2}}$ exists for all $\text{x}\in(-3,4)$
So, f(x) is differentiable on (-3, 4).
Thus, both the conditions of Lagrange's theorem are satisfied.
Consequently, there exist some $\text{c}\in(-3,4)$
such that$\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(-3)}{4+3}=\frac{\text{f}(4)-\text{f}(-3)}{7}$
Now, $\text{f}(\text{x})=\sqrt{25-\text{x}^2}$
$\text{f}'(\text{x})=\frac{-\text{x}}{\sqrt{25-\text{x}^2}},\text{f}(-3)=4,\text{f}(4)=3$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(4)-\text{f}(-3)}{4+3}$
$\Rightarrow\frac{-\text{x}}{\sqrt{25-\text{x}^2}}=\frac{3-4}{7}$
$\Rightarrow49\text{x}^2=25-\text{x}^2$
$\Rightarrow\text{x}=\pm\frac{1}{\sqrt2}$
Thus, $\text{c}=\pm\frac{1}{\sqrt2}\in(-3,4)$ such that $\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(-3)}{4-(-3)}.$
Hence, Lagrange's mean value theorem is verified.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\int\text{x}\tan^2\text{x dx}$