00:01
In this question, we are given a function f of x is equal to x squared times e raised the part 8x.
00:12
And we are given in towers negative of infinity to a, a to b and b to infinity.
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Such that a and b are critical points of f of x.
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We need to obtain, for the first subpart, the values of a and b.
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That means we need to obtain the critical points of f of x.
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For that, let us first differentiate the function.
00:48
We get f -dash -x is equal to d over dx of x squared times e -raise to par a, now using the product rule differentiating we get x squared times d over dx of e raise to par 8 x plus e raise to per 8 x hence d over d x of x squared.
01:12
This gives x squared times 8 times e r r r s to per 8 x plus e raise to per 8 x times 2x.
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Now, factor out 2x e -raise to par 8x.
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We are left with 4x plus 1.
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Therefore, this is the value of f -h of x.
01:39
Now, we know that at critical points, the value of the derivative is 0.
01:48
Equating it with 0, we get 0 is equal to 2x times e -race to par 8x times 4x plus 1.
01:57
Equating both the factors with 0, we get x is equal to 0 because e -raise to per rate x cannot be equal to 0.
02:10
Also, 4x plus 1 is equal to 0 implies x is equal to negative of 1 over 4.
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Therefore, the critical points are x is equal to 0 and x is equal to negative of 1 over 4.
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Therefore, the value of a is negative of 1 over 4 because it is smaller and the value of b is 0.
02:34
Therefore, these are the solutions for the first subpart.
02:39
Further, for the second subpart, we need to check that in the given intervals, negative infinity to a, a to b and b to infinity, whether the function is increasing or decreasing.
02:58
Now first of all, these intervals can now be written as negative infinity to negative of 1 over 4, then negative of 1 over 4 to 0, then 0 to infinity.
03:10
Now to check whether the function are increasing or decreasing in these intervals, again differentiate the given function, that is, f double dash of x is equal to d over d x of 2 times of x, e -raise to per 8x times 4x plus 1.
03:34
Now, let us take this at first term and this is a second term by using product rule we get 2 times x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x 3x plus 1 .4x plus 1 times d over d x x x x x 3 times d over d x x x x x x x x 3 times x x x x x x x x x x 3x 4x 4x 4x...