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Utilizes the second derivative test to classify the critical points of the function,
h ( x ) = 3x5 - 5x3 + 3
Solution
The second derivative is,
h′′ ( x ) + 60x3 - 30x
The three critical points (x = -1, x = 0, & x = 1) of this function are all critical points where the first derivative is zero therefore we know that at least we have a possibility that the Second Derivative Test will work. For each of these the value of the second derivative is,
h′′ ( -1) = -30 h′′ (0) = 0 h′′ (1) = 30
The second derivative at x = -1 is -ve the second derivative at x = 1 is +ve and therefore we have a relative minimum here by the Second Derivative Test
In the case of x = 0 the second derivative is zero and therefore we can't utilizes the Second Derivative Test to classify this critical point.
ABC is a right triangle right-angled at C and AC=√3 BC. Prove that ∠ABC=60 o . Ans: Tan B = AC/BC Tan B = √3 BC/BC Tan B =√3 ⇒ Tan B = Tan 60 ⇒ B = 60
Solve 4 cos(t )= 3 on[-8,10]. Solution : Here the first step is identical to the problems in the previous section. First we need to isolate the cosine on one side by itself & t
4. Two hosts, one on East (host A) and one on the west coast (host B) of the USA are exchanging data. Suppose A is sending a large file to B. The file is split into packets of size
Every point (x,y) on the curve y=log2 3x is transferred to a new point by the following translation (x',y')=(x+m,y+n), where m and n are integers. The set of (x',y') form the curve
how to solve
Proof of: lim q →0 sin q / q = 1 This proofs of given limit uses the Squeeze Theorem. Though, getting things set up to utilize the Squeeze Theorem can be a somewha
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log base 5 (3-2x) + log base 5 (2+x) = 1
Which number below is described by the following statements? The hundredths digit is 4 and the tenths digit is twice the thousandths digit. a. 0.643 b. 0.0844 c. 0.446 d. 0.0142
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