For the curve $y=x^2$, take two points with $x$-coordinates $a$ and $b$, where \(a

Question

For the curve $y=x^2$, take two points with $x$-coordinates $a$ and $b$, where $a<b$. Show that the secant gradient between these points is $a+b$. Show that this equals the tangent gradient at the midpoint $x=\dfrac{a+b}{2}$. A student concludes: “For every curve, the secant gradient equals the tangent gradient at the midpoint.” Test this claim using $y=x^3$ between $x=0$ and $x=2$.

Accepted Answer

(a) Secant gradient $a+b$, (b) Equal to the quadratic’s midpoint gradient; (c) Cubic counterexample: $4\ne3$.

Differentiation

Differentiation

Solution:

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