If the derivative of the function $$f(x)=ax2+b$$, x

Question

If the derivative of the function $$f(x)=ax2+b$$,    x<−$$1bx2+ax+4$$,    x≥−$$1$$ is everywhere continuous, then

Infinity Learn · Accepted Answer

$$f(x)=ax2+b$$,              x<−$$1bx2+ax+4$$,  x≥−$$1$$⇒f'$$x=2ax$$,            x<−$$12bx+a$$,    x≥−$$1Since$$ $$f(x)$$ is differentiable at $$x=$$−$$1$$,∴fx is continuous at $$x=$$−$$1$$∴limx→$$1$$−$$f(x)=limx$$→$$1+f(x)$$⇒$$a+b=b$$−$$a+4$$⇒$$a=2$$             ....(1)Also$$, since f'$$(x) is continuous, therefore, f'(x) is continuous at $$x=$$−$$1$$.∴limx→−$$1$$−f'(x)=limx$$→−$$1+f$$'$$(x)⇒−$$2a=$$−$$2b+a$$                          ....(2)Using$$ $$(1) in (2), we get, $$b=3$$∴$$a=2$$, $$b=3$$.

$If the derivative of the function f (x) = \{\begin{cases} a x^{2} + b, x < - 1 \\ b x^{2} + a x + 4, x \geq - 1 \end{cases} is everywhere continuous, then$

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If the derivative of the function f(x)=ax2+b, x<−1bx2+ax+4, x≥−1 is everywhere continuous, then

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$If the derivative of the function f (x) = \{\begin{cases} a x^{2} + b, x < - 1 \\ b x^{2} + a x + 4, x \geq - 1 \end{cases} is everywhere continuous, then$