As per the given question,
Show that log x/x has a maximum value at x = e.
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Exercise 18.3
The function y = a log x + bx^2 + x has extreme values at x = 1 and x = 2. Find a and b.
Given \[y\text{ }=\text{ }a\text{ }log\text{ }x\text{ }+\text{ }b{{x}^{2}}~+\text{ }x\] On differentiating we get Given that extreme values exist at \[x\text{ }=\text{ }1,\text{ }2\] \[a\text{...
Find the local extremum values of the following functions: f (x) = – (x – 1)^3(x + 1)^2
Given $f(x)=-(x-1)^{3}(x+1)^{2}$ $ \begin{array}{l} f^{\prime}(x)=-3(x-1)^{2}(x+1)^{2}-2(x-1)^{3}(x+1) \\ =-(x-1)^{2}(x+1)(3 x+3+2 x-2) \\ =-(x-1)^{2}(x+1)(5 x+1) \\ f^{\prime...
Find the local extremum values of the following functions: (i) f(x) = (x – 1) (x – 2)^2
Solution: (i) $ \begin{array}{l} \text { Given } f(x)=(x-1)(x-2)^{2} \\ f(x)=(x-2)^{2}+2(x-1)(x-2) \\ =(x-2)(x-2+2 x-2) \\ =(x-2)(3 x-4) \\ f^{\prime}(x)=(3 x-4)+3(x-2) \end{array} $ For maxima and...
Find the points of local maxima or local minima and corresponding local maximum and local minimum values of each of the following functions. Also, find the points of inflection, if any: f (x) = x e^x
Given $\mathrm{f}(\mathrm{x})=\mathrm{x} \mathrm{e}^{x}$ $ \begin{array}{l} f(x)=e^{x}+x e^{i}=e^{x}(x+1) \\ f^{\prime}(x)=e^{2}(x+1)+e^{t} \\ =e^{2}(x+2) \end{array} $ For maxima and minima, $...
Find the points of local maxima or local minima and corresponding local maximum and local minimum values of each of the following functions. Also, find the points of inflection, if any: (i) f(x) = (x – 1) (x + 2)^2 (ii) f (x) = 2/x – 2/x^2, x > 0
(i) Given $f(x)=(x-1)(x+2)^{2}$ $ \begin{array}{l} \therefore \mathrm{f}(\mathrm{x})=(\mathrm{x}+2)^{2}+2(\mathrm{x}-1)(\mathrm{x}+2) \\ =(\mathrm{x}+2)(\mathrm{x}+2+2 \mathrm{x}-2) \\...
Find the points of local maxima or local minima and corresponding local maximum and local minimum values of each of the following functions. Also, find the points of inflection, if any: (i) f(x) = x^4 – 62x^2 + 120x + 9 (ii) f (x) = x^3 – 6x^2 + 9x + 15
(i) Given $f(x)=x^{4}-62 x^{2+} 120 x+9$ $ \begin{array}{l} \therefore \mathrm{f}(\mathrm{x})=4 \mathrm{x}^{\mathrm{a}}-124 x+120=4\left(\mathrm{x}^{a}-31 \mathrm{x}+30\right) \\...