Question

please solve this question .​

Answer 1

refer to pic Above please

Answer 2

f(x)=( - 3/4) x^4−8x^3- (45/2)x^2+105

f′(x)= - 3x^3 - 24x^2 - 45x

=−3x[x^2 +8x+15]

=−3x[x^2 +5x+3x+15]

=−3x[x(x+5)+3(x+5)]

=−3x(x+5)(x+3)

Now for critical points f′ (x)=0 gives

x=−5 and x=−3

∴at x=−5 is a point of minima

∴at x=−3 is a point of maxima.

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