Prove that the following functions do not have maxima or minima: (i) f(x) = e^ x (ii) g(x) = logx (iii) h(x) = x ^3 + x^ 2 + x + 1
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(i) f(x) = e^x
differentiate with respect to x,
f'(x) = e^x
Now, if f’(x) = 0, then e^x = 0.
But, the exponential function can never assume 0 for any value of x.
Therefore, there does not exist c ∈ R such that
f’(c) = 0
Hence, function f(x) does not have maxima or minima.
(ii) g(x) = logx
differentiate with respect to x,
g'(x) = 1/x
now , if f'(x) = 0 then, 1/x = 0
but , we know there are no any value of x in which 1/x attains zero.
hence, function g(x) doesn't have maxima or minima.
(iii) h(x) = x³ + x² + x + 1
differentiate with respect to x,
h'(x) = 3x² + 2x + 1
now, h'(x) = 0 , then, 3x² + 2x + 1 = 0
but here, D = b² - 4ac = 2² - 4.3 < 0
so, 3x² + 2x + 1 ≠ 0 for all real numbers of x
therefore, function h(x) doesn't have maxima or minima.
differentiate with respect to x,
f'(x) = e^x
Now, if f’(x) = 0, then e^x = 0.
But, the exponential function can never assume 0 for any value of x.
Therefore, there does not exist c ∈ R such that
f’(c) = 0
Hence, function f(x) does not have maxima or minima.
(ii) g(x) = logx
differentiate with respect to x,
g'(x) = 1/x
now , if f'(x) = 0 then, 1/x = 0
but , we know there are no any value of x in which 1/x attains zero.
hence, function g(x) doesn't have maxima or minima.
(iii) h(x) = x³ + x² + x + 1
differentiate with respect to x,
h'(x) = 3x² + 2x + 1
now, h'(x) = 0 , then, 3x² + 2x + 1 = 0
but here, D = b² - 4ac = 2² - 4.3 < 0
so, 3x² + 2x + 1 ≠ 0 for all real numbers of x
therefore, function h(x) doesn't have maxima or minima.
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