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Find the points on the curve y=3x^4+2x^3-1 at which the tangents is parallel to the x-axis.(a) (0,1) and \((-\frac{1}{2},-\frac{15}{16})\)(b) (0,-1) and \((-\frac{1}{2},-\frac{15}{16})\)(c) (0,-1) and \((\frac{1}{2},-\frac{15}{16})\)(d) (0,1) and \((\frac{1}{2},\frac{15}{16})\)This question was addressed to me in my homework.The origin of the question is Derivatives Application topic in section Application of Derivatives of Mathematics – Class 12

Answer»

Right OPTION is (b) (0,-1) and \((-\frac{1}{2},-\frac{15}{16})\)

EXPLANATION: GIVEN that, y=3x^4+2x^3-1

Differentiating w.r.t x, we GET

\(\frac{dy}{dx}\)=12x^3+6x^2

The tangent is parallel to x-axis, which implies that the slope \(\frac{dy}{dx}\) is 0.

∴12x^3+6x^2=0

6x^2 (2x+1)=0

⇒x=0,-\(\frac{1}{2}\)

If x=0

y=3(0)+2(0)-1=-1

If x=-\(\frac{1}{2}\)

y=3\((-\frac{1}{2})^4+2(-\frac{1}{2})^3-1\)

y=\(\frac{3}{16}-\frac{2}{8}-1\)

y=-\(\frac{15}{16}\)

Hence, the points at which the tangent is parallel to the x-axis is (0,-1) and \((-\frac{1}{2},-\frac{15}{16})\).


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