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Question
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
Solution
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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