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Worked Examples · Example 19

Q.Find the equation of the tangent and normal to the curve f(x)=ex+x2+1f(x) = e^x + x^2 + 1 at the point (0, 2) on it.

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✓ Free question

Differentiate, evaluate the slope at (0,2)(0,2), then use point-slope for the tangent and the negative reciprocal slope for the normal.

Slope of tangent: mt=f′(x0)m_t=f'(x_0). Slope of normal: mn=−1f′(x0)m_n=-\dfrac{1}{f'(x_0)}. Line through (x0,y0)(x_0,y_0): y−y0=m(x−x0)y-y_0=m(x-x_0).

  • f′(x0)f'(x_0) = derivative of the curve at the point of contact.
  1. Differentiate f(x)=ex+x2+1f(x)=e^x+x^2+1:

f′(x)=ex+2x.f'(x)=e^x+2x.

  1. Evaluate at x=0x=0:

mt=f′(0)=e0+2(0)=1.m_t=f'(0)=e^0+2(0)=1.

  1. Tangent through (0,2)(0,2) with slope 11:

y−2=1(x−0) ⇒ y=x+2.y-2=1(x-0)\ \Rightarrow\ y=x+2.

  1. Normal slope mn=−11=−1m_n=-\dfrac{1}{1}=-1; line through (0,2)(0,2):

y−2=−1(x−0) ⇒ x+y=2.y-2=-1(x-0)\ \Rightarrow\ x+y=2.

✓Final answer

Tangent: y=x+2y=x+2 (i.e. x−y+2=0x-y+2=0). Normal: x+y=2x+y=2.

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