Application of Derivatives Test

Result

Application of Derivatives Test - 7

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Question 1
1 / -0

Tangents to the curve y= x³ + 3x at x = -1 and x = 1 are

A
parallel

B
intersecting at an angle of 45^o.

C
intersecting at right angles

D
intersecting obliquely but not at an angle of 45^o

Solution

Given y =x³ + 3x

⇒ dy/dx = 3x²+ 3

Slope of tangent at x = 1 = 6 and

Slope of tangent at x = -1 = 6

Hence, the two tangents are parallel.
Question 2
1 / -0

The smallest value of the polynomial x³ - 18x² + 96x in the interval [0, 9] is

A
0

B
160

C
135

D
128

Solution

Let f(x) = x³ - 18x²+ 96x

⇒⇒ f'(x) = 3x² - 36x + 96 = 3[x² - 12x +32] = 3(x - 4)(x - 8)

For maximum and minimum values of x, we have f'(x) = 0

⇒ 3(x - 4)(x - 8) ⇒ x = 4, 8

Both of these values lies in the given interval [0,9]

Now, f''(x) = 6x - 36

When x = 8 we get f''(x) = 12 > 0

Since at x = 8, f'(x)= 0 and f''(x)>0, we get f(x) is minimum at x = 8 in (0,9)

Now, we have to find minimum values at the end points of the given interval

We have, f(0) = 0 and f(9) = 135

Hence, the minimum value of f(x) = 0 at x= 0 in [0,9]