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Show that there lies a point on the curve f(x)=x(x+3)eπ2,-3≤x≤0 where tangent drawn is parallel to the x-axis - Mathematics

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प्रश्न

Show that there lies a point on the curve `f(x) = x(x + 3)e^(pi/2), -3 ≤ x ≤ 0` where tangent drawn is parallel to the x-axis

योग

उत्तर

f(x) = `x(x + 3)e^(pi/2), -3 ≤ x ≤ 0`

f(x) is continuous in [– 3, 0]

f(x) is differentiable in (– 3, 0)

f(– 3) = `-3(-3 + 3)"e"^((-pi)/2)` = 0

f(0) = 0

⇒ f(– 3) = f(0) = 0

Since the tangent is parallel to x-axis.

f'(c) = 0

`"e"^((- pi)/2) (2"c" + 3) = 0 `

Where f'(x) = `"e"^((- pi)/2) (2x + 3)`

2c + 3 = 0

c = `- 3/2 ∈ (-3, 0)`

∴ The point lies on the curve.

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Mean Value Theorem
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Applications of Differential Calculus - Exercise 7.3 [पृष्ठ २२]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 12 TN Board
अध्याय 7 Applications of Differential Calculus
Exercise 7.3 | Q 9 | पृष्ठ २२

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